**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Turtle and Inversions (Hard Version)
time_limit_ms: 2000
memory_limit_mb: 512
Description: This is a hard version of this problem. The differences between the versions are the constraint on $m$ and $r_i < l_{i + 1}$ holds for each $i$ from $1$ to $m - 1$ in the easy version. You can make hacks only if both versions of the problem are solved.

Turtle gives you $m$ intervals $[l_1, r_1], [l_2, r_2], \ldots, [l_m, r_m]$. He thinks that a permutation $p$ is interesting if there exists an integer $k_i$ for every interval ($l_i \le k_i < r_i$), and if he lets $a_i = \max\limits_{j = l_i}^{k_i} p_j, b_i = \min\limits_{j = k_i + 1}^{r_i} p_j$ for every integer $i$ from $1$ to $m$, the following condition holds:

$$\max\limits_{i = 1}^m a_i < \min\limits_{i = 1}^m b_i$$

Turtle wants you to calculate the maximum number of inversions of all interesting permutations of length $n$, or tell him if there is no interesting permutation.

An inversion of a permutation $p$ is a pair of integers $(i, j)$ ($1 \le i < j \le n$) such that $p_i > p_j$.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^3$). The description of the test cases follows.

The first line of each test case contains two integers $n, m$ ($2 \le n \le 5 \cdot 10^3, 0 \le m \le 5 \cdot 10^3$) — the length of the permutation and the number of intervals.

The $i$-th of the following $m$ lines contains two integers $l_i, r_i$ ($1 \le l_i < r_i \le n$) — the $i$-th interval. Note that there may exist identical intervals (i.e., there may exist two different indices $i, j$ such that $l_i = l_j$ and $r_i = r_j$).

It is guaranteed that the sum of $n$ over all test cases does not exceed $5 \cdot 10^3$ and the sum of $m$ over all test cases does not exceed $5 \cdot 10^3$.

Output: For each test case, if there is no interesting permutation, output a single integer $-1$.

Otherwise, output a single integer — the maximum number of inversions.

Examples:
input:
8
2 0
2 1
1 2
5 1
2 4
8 3
1 4
2 5
7 8
7 2
1 4
4 7
7 3
1 2
1 7
3 7
7 4
1 3
4 7
1 3
4 7
7 3
1 2
3 4
5 6
output:
1
0
8
18
-1
-1
15
15

Note: In the third test case, the interesting permutation with the maximum number of inversions is $[5, 2, 4, 3, 1]$.

In the fourth test case, the interesting permutation with the maximum number of inversions is $[4, 3, 8, 7, 6, 2, 1, 5]$. In this case, we can let $[k_1, k_2, k_3] = [2, 2, 7]$.

In the fifth and the sixth test case, it can be proven that there is no interesting permutation.

In the seventh test case, the interesting permutation with the maximum number of inversions is $[4, 7, 6, 3, 2, 1, 5]$. In this case, we can let $[k_1, k_2, k_3, k_4] = [1, 6, 1, 6]$.

In the eighth test case, the interesting permutation with the maximum number of inversions is $[4, 7, 3, 6, 2, 5, 1]$.

Buggy Code:
#include<bits/stdc++.h>
#define cir(i,a,b) for(int i=a;i<b;++i)
using namespace std;
static constexpr auto _inf=(int)(1e9+7);
int main(){
    ios::sync_with_stdio(false),cin.tie(nullptr);
    int T;cin>>T;
    while(T--) [&](){
        int n,q;cin>>n>>q;
        auto check=[&](vector<pair<int,int>> a,vector<int> val){
            ranges::sort(a);
            const auto nx=(int)(a.size());
            auto cur=0;
            vector f(2,vector(n+1,-_inf));
            f[cur][0]=0;
            for(auto&[l,r]:a) if(l<r+1){
                cur^=1;
                ranges::fill(f[cur],-_inf);
                if(!is_sorted(val.begin()+l,val.begin()+r+1)) return -1;
                const auto c0=count(val.begin()+l,val.begin()+r+1,0),c1=count(val.begin()+l,val.begin()+r+1,1);
                cir(las,0,l+1) cir(w,c1,r-l+1-c0+1){
                    f[cur][las+w]=max(f[cur][las+w],f[cur^1][las]+las*(r-l+1-w));
                }
                clog.flush();
            }
            auto res=0;
            cir(i,0,n) res=max(res,f[cur][i]+i*(i-1)/2+(n-i)*(n-i-1)/2);
            return res;
        };
        vector<pair<int,int>> a(q);
        vector<int> ok(n),sw(n,-1);
        for(auto&[l,r]:a){
            cin>>l>>r;--l;--r;
        }
        ranges::sort(a);
        for(auto&[l,r]:a){
            if(sw[l]>-1&&sw[l]!=0) return println("-1");
            if(sw[r]>-1&&sw[r]!=1) return println("-1");
            sw[l]=0;sw[r]=1;
            ++l;--r;
        }
        auto vaildmg=[&](pair<int,int> a,pair<int,int> b){
            return !(a.second<b.first||b.second<a.first);
        };
        auto getmg=[&](pair<int,int> a,pair<int,int> b){
            if(a.first>b.first) swap(a,b);
            return pair(max(a.first,b.first),min(a.second,b.second));
        };
        vector<pair<int,int>> rg;
        for(auto&[l,r]:a) if([&](){
            for(auto&[al,ar]:rg){
                if(!vaildmg({l,r},{al,ar})) continue;
                const auto[ml,mr]=getmg({l,r},{al,ar});
                cir(i,min(l,al),ml){
                    if(sw[i]>-1&&sw[i]!=0) return println("-1"),true;
                    sw[i]=0;
                }
                cir(i,mr+1,max(r,ar)+1){
                    if(sw[i]>-1&&sw[i]!=1) return println("-1"),true;
                    sw[i]=1;
                }
                tie(al,ar)=tie(ml,mr);
                return false;
            }
            rg.emplace_back(l,r);
            return false;
        }()) return;
        for(auto&[l,r]:rg) cir(i,l,r+1) ok[i]=true;
        cir(i,0,n) if(!ok[i]) rg.emplace_back(i,i);
        println("{}",check(rg,sw));
    }();
    return 0;
}

**Instructions:**
- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.
- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.
- Output the test input directly, not code to generate it.

**Output format:**
```plaintext
{test input}
```

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided buggy code, we need to focus on edge cases and scenarios that maximize the failure probability. The code has several potential issues, including handling of intervals, sorting, and the logic for checking the validity of permutations. Let's consider the following steps:

1. **Edge Cases for Intervals**:
   - Intervals that overlap or are adjacent.
   - Intervals that cover the entire permutation.
   - Intervals that are very small or very large.

2. **Invalid Permutations**:
   - Cases where no valid permutation can satisfy the conditions.
   - Cases where the intervals are such that the conditions for \(a_i\) and \(b_i\) cannot be met.

3. **Maximum Inversions**:
   - Cases where the permutation with the maximum number of inversions is required.

Given these considerations, let's construct a small-scale test input that includes overlapping intervals and a scenario where no valid permutation exists.

### Test Input
```plaintext
3
5 2
1 3
3 5
8 3
1 4
2 5
7 8
7 3
1 2
3 4
5 6
```

### Explanation
1. **Test Case 1**:
   - Intervals: \([1, 3]\) and \([3, 5]\)
   - Overlapping intervals. The code should handle this correctly and find the maximum number of inversions.

2. **Test Case 2**:
   - Intervals: \([1, 4]\), \([2, 5]\), and \([7, 8]\)
   - Overlapping and non-overlapping intervals. This should test the code's ability to handle multiple intervals and ensure the conditions for \(a_i\) and \(b_i\) are met.

3. **Test Case 3**:
   - Intervals: \([1, 2]\), \([3, 4]\), and \([5, 6]\)
   - Non-overlapping intervals. This should test the code's ability to handle disjoint intervals and ensure the conditions for \(a_i\) and \(b_i\) are met.

This test input should help expose any bugs in the code, particularly in handling overlapping intervals and ensuring the validity of permutations.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Turtle and Inversions (Hard Version)
time_limit_ms: 2000
memory_limit_mb: 512
Description: This is a hard version of this problem. The differences between the versions are the constraint on $m$ and $r_i < l_{i + 1}$ holds for each $i$ from $1$ to $m - 1$ in the easy version. You can make hacks only if both versions of the problem are solved.

Turtle gives you $m$ intervals $[l_1, r_1], [l_2, r_2], \ldots, [l_m, r_m]$. He thinks that a permutation $p$ is interesting if there exists an integer $k_i$ for every interval ($l_i \le k_i < r_i$), and if he lets $a_i = \max\limits_{j = l_i}^{k_i} p_j, b_i = \min\limits_{j = k_i + 1}^{r_i} p_j$ for every integer $i$ from $1$ to $m$, the following condition holds:

$$\max\limits_{i = 1}^m a_i < \min\limits_{i = 1}^m b_i$$

Turtle wants you to calculate the maximum number of inversions of all interesting permutations of length $n$, or tell him if there is no interesting permutation.

An inversion of a permutation $p$ is a pair of integers $(i, j)$ ($1 \le i < j \le n$) such that $p_i > p_j$.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^3$). The description of the test cases follows.

The first line of each test case contains two integers $n, m$ ($2 \le n \le 5 \cdot 10^3, 0 \le m \le 5 \cdot 10^3$) — the length of the permutation and the number of intervals.

The $i$-th of the following $m$ lines contains two integers $l_i, r_i$ ($1 \le l_i < r_i \le n$) — the $i$-th interval. Note that there may exist identical intervals (i.e., there may exist two different indices $i, j$ such that $l_i = l_j$ and $r_i = r_j$).

It is guaranteed that the sum of $n$ over all test cases does not exceed $5 \cdot 10^3$ and the sum of $m$ over all test cases does not exceed $5 \cdot 10^3$.

Output: For each test case, if there is no interesting permutation, output a single integer $-1$.

Otherwise, output a single integer — the maximum number of inversions.

Examples:
input:
8
2 0
2 1
1 2
5 1
2 4
8 3
1 4
2 5
7 8
7 2
1 4
4 7
7 3
1 2
1 7
3 7
7 4
1 3
4 7
1 3
4 7
7 3
1 2
3 4
5 6
output:
1
0
8
18
-1
-1
15
15

Note: In the third test case, the interesting permutation with the maximum number of inversions is $[5, 2, 4, 3, 1]$.

In the fourth test case, the interesting permutation with the maximum number of inversions is $[4, 3, 8, 7, 6, 2, 1, 5]$. In this case, we can let $[k_1, k_2, k_3] = [2, 2, 7]$.

In the fifth and the sixth test case, it can be proven that there is no interesting permutation.

In the seventh test case, the interesting permutation with the maximum number of inversions is $[4, 7, 6, 3, 2, 1, 5]$. In this case, we can let $[k_1, k_2, k_3, k_4] = [1, 6, 1, 6]$.

In the eighth test case, the interesting permutation with the maximum number of inversions is $[4, 7, 3, 6, 2, 5, 1]$.

Buggy Code:
#include <bits/stdc++.h>
#define l first
#define r second
using namespace std;
typedef long long ll;
const int inf = 1e9;
const ll INF = 1e15;
const int N = 4e5;
inline ll read() {
	ll s = 0,f = 1;char ch = getchar();
	while (!isdigit(ch)) f = ch == '-' ? -1 : 1, ch = getchar();
	while (isdigit(ch)) s = (s << 3) + (s << 1) + ch - '0', ch = getchar();
	return s*f;
}
const int mod = 998244353;
int getmod(int x) {
	return x - (x >= mod) * mod;
}
int qpow(int a,int b) {
	int res = 1;
	while (b) {
		if (b & 1) res = 1ll * res * a % mod;
		b >>= 1, a = 1ll * a * a % mod;
	}
	return res;
}
int n,m,ct[N + 10],ct2[N + 10];
pair<int,int> a[N + 10];
pair<int,int> b[N + 10];
int main() {
	int T = read();
	while (T -- ) {
		n = read(), m = read();
		for (int i = 1;i <= m;i ++ ) {
			int l = read(),r = read();
			a[i] = {l,r};
		}
		sort(a + 1,a + m + 1,[](pair<int,int> x,pair<int,int> y) {
			return x.l != y.l ? x.l < y.l : x.r > y.r;
		});
		for (int i = 1;i <= m;i ++ ) {
			b[i] = a[i];
		}
		int ans = 0,nn = 0;
		for (int i = 1;i <= m;i ++ )
			for (int j = 1;j <= m;j ++ ) {
				if (a[j].l >= a[i].l && a[j].l <= a[i].r) b[i].l = max(b[i].l,a[j].l);
				if (a[j].r >= a[i].l && a[j].r <= a[i].r) b[i].r = min(b[i].r,a[j].r);
			}
		ct[m + 1] = ct2[m + 1] = 0;
		for (int i = m;i;i -- ) {
			b[i].r --;
			if (b[i].l > b[i].r) {
				puts("-1");
				goto bre;
			}
//			ct[i] = ct[i + 1] + a[i].r - b[i].r;
//			ct2[i] = ct2[i + 1] + b[i].l - a[i].l + 1;
//			cout << "b:" << i << ' ' << a[i].l << ' ' << a[i].r << ' ' << b[i].l << ' ' << b[i].r << endl;
		}
		a[0].r = -1;
		for (int i = 1;i <= m;i ++ ) {
			if (a[i].r > a[i - 1].r) a[++nn] = a[i], b[nn] = b[i];
		}
		m = nn;
		for (int j = 1;j <= m;j ++ ) {
			int l = a[j].l,r = a[j].r;
			int tr = j - 1;
			while (tr < m && a[tr + 1].l <= r) tr ++;
			ct[j] = a[tr].r - b[j].r;
			ct2[j] = b[j].l - a[j].l + 1;
//			cout << "ct:" << j << ' ' << ct[j] << ' ' << ct2[j] << endl;
//			for (int k = j;k <= tr;k ++ ) ct[j] += a[k].r - b[j].r, ct2[j] +=
			j = tr;
		}
		for (int j = m;j;j -- ) ct[j] += ct[j + 1], ct2[j] += ct2[j + 1];
		for (int i = 1;i <= n;i ++ ) {
			//<i >= i
			int p = 0,now = n,now2 = i - 1,res = 0;
			if (now - i + 1 < ct[1] || now2 < ct2[1]) continue;
			for (int j = 1;j <= m;j ++ ) {
				int l = a[j].l,r = a[j].r;
				for (int k = p + 1;k < l;k ++ )
					if (now - i + 1 > ct[j]) res += now2 + now - i, now --;
					else res += now2 - 1, now2 --;
				int tr = j - 1;
				while (tr < m && a[tr + 1].l <= r) tr ++;
				int mx = min(a[tr].r - b[j].l,now - i + 1 - ct[tr + 1]);
//				cout << "!!!" << j << ' ' << now << ' ' << mx << endl;
				for (int k = l;k <= a[tr].r - mx;k ++ ) res += now2 - 1, now2 --;
				for (int k = a[tr].r - mx + 1;k <= a[tr].r;k ++ ) res += now2 + now - i, now --;
				p = a[tr].r;
				j = tr;
			}
			while (p < n) {
				if (now >= i) res += now2 + now - i, now --;
				else res += now2 - 1, now2 --;
				p ++;
			}
//			cout << "!!" << i << ' ' << res << endl;
			ans = max(ans,res);
		}
		cout << ans << endl;
		bre:;
	}
	return 0;
}

**Instructions:**
- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.
- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.
- Output the test input directly, not code to generate it.

**Output format:**
```plaintext
{test input}
```

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided code, we need to focus on edge cases and scenarios that maximize the failure probability. The code is complex, and the issue might be related to how intervals are handled, particularly in edge cases where intervals overlap or are very close to each other.

Let's consider a few key points:
1. **Overlapping Intervals**: Intervals that overlap significantly can cause issues in the logic that handles the merging of intervals.
2. **Single Intervals**: Intervals that cover the entire range of the permutation can simplify the problem and test the boundary conditions.
3. **No Intervals**: Test cases with no intervals should return the maximum number of inversions for a permutation of length \( n \).
4. **Edge Cases**: Intervals that start or end at the boundaries of the permutation can be tricky.

Given these points, let's create a small-scale test input that covers these scenarios:

```plaintext
4
5 1
1 5
5 2
1 3
3 5
5 3
1 2
2 4
4 5
5 0
```

### Explanation:
1. **Test Case 1**: \( n = 5, m = 1 \)
   - Interval: \([1, 5]\)
   - This interval covers the entire permutation, so it should return the maximum number of inversions for a permutation of length 5, which is \( 10 \).

2. **Test Case 2**: \( n = 5, m = 2 \)
   - Intervals: \([1, 3]\), \([3, 5]\)
   - These intervals overlap, and the code should handle the merging of these intervals correctly.

3. **Test Case 3**: \( n = 5, m = 3 \)
   - Intervals: \([1, 2]\), \([2, 4]\), \([4, 5]\)
   - These intervals are closely spaced and should test the logic for handling multiple intervals.

4. **Test Case 4**: \( n = 5, m = 0 \)
   - No intervals.
   - This should return the maximum number of inversions for a permutation of length 5, which is \( 10 \).

This test input is designed to cover a range of scenarios that can help expose potential bugs in the provided code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Turtle and Inversions (Hard Version)
time_limit_ms: 2000
memory_limit_mb: 512
Description: This is a hard version of this problem. The differences between the versions are the constraint on $m$ and $r_i < l_{i + 1}$ holds for each $i$ from $1$ to $m - 1$ in the easy version. You can make hacks only if both versions of the problem are solved.

Turtle gives you $m$ intervals $[l_1, r_1], [l_2, r_2], \ldots, [l_m, r_m]$. He thinks that a permutation $p$ is interesting if there exists an integer $k_i$ for every interval ($l_i \le k_i < r_i$), and if he lets $a_i = \max\limits_{j = l_i}^{k_i} p_j, b_i = \min\limits_{j = k_i + 1}^{r_i} p_j$ for every integer $i$ from $1$ to $m$, the following condition holds:

$$\max\limits_{i = 1}^m a_i < \min\limits_{i = 1}^m b_i$$

Turtle wants you to calculate the maximum number of inversions of all interesting permutations of length $n$, or tell him if there is no interesting permutation.

An inversion of a permutation $p$ is a pair of integers $(i, j)$ ($1 \le i < j \le n$) such that $p_i > p_j$.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^3$). The description of the test cases follows.

The first line of each test case contains two integers $n, m$ ($2 \le n \le 5 \cdot 10^3, 0 \le m \le 5 \cdot 10^3$) — the length of the permutation and the number of intervals.

The $i$-th of the following $m$ lines contains two integers $l_i, r_i$ ($1 \le l_i < r_i \le n$) — the $i$-th interval. Note that there may exist identical intervals (i.e., there may exist two different indices $i, j$ such that $l_i = l_j$ and $r_i = r_j$).

It is guaranteed that the sum of $n$ over all test cases does not exceed $5 \cdot 10^3$ and the sum of $m$ over all test cases does not exceed $5 \cdot 10^3$.

Output: For each test case, if there is no interesting permutation, output a single integer $-1$.

Otherwise, output a single integer — the maximum number of inversions.

Examples:
input:
8
2 0
2 1
1 2
5 1
2 4
8 3
1 4
2 5
7 8
7 2
1 4
4 7
7 3
1 2
1 7
3 7
7 4
1 3
4 7
1 3
4 7
7 3
1 2
3 4
5 6
output:
1
0
8
18
-1
-1
15
15

Note: In the third test case, the interesting permutation with the maximum number of inversions is $[5, 2, 4, 3, 1]$.

In the fourth test case, the interesting permutation with the maximum number of inversions is $[4, 3, 8, 7, 6, 2, 1, 5]$. In this case, we can let $[k_1, k_2, k_3] = [2, 2, 7]$.

In the fifth and the sixth test case, it can be proven that there is no interesting permutation.

In the seventh test case, the interesting permutation with the maximum number of inversions is $[4, 7, 6, 3, 2, 1, 5]$. In this case, we can let $[k_1, k_2, k_3, k_4] = [1, 6, 1, 6]$.

In the eighth test case, the interesting permutation with the maximum number of inversions is $[4, 7, 3, 6, 2, 5, 1]$.

Buggy Code:
//ayanami���ӣ���籣�ӣ����豣�ӣ�MDR���ӣ�ﱵ��ֱ��ӣ�M99���ӣ��˵�����
#include<bits/stdc++.h>
using namespace std;
int p1=1000000,p2=0;
char buf[1000005],wb[1000005];
int gc() {
	if(p1>=1000000)fread(buf,1,1000000,stdin),p1=0;
	return buf[p1++];
}
#define gc getchar
#define Loli true
#define Kon xor true
long long getint() {
	long long ret=0,flag=1;
	char c=gc();
	while(c<'0'||c>'9') {
		if(c=='-')flag=-1;
		c=gc();
	}
	while(c<='9'&&c>='0') {
		ret=(ret<<3)+(ret<<1)+c-'0';
		c=gc();
	}
	return ret*flag;
}
void pc(char x) {
	if(p2>=1000000)fwrite(wb,1,1000000,stdout),p2=0;
	wb[p2++]=x;
}
void wrt(long long x) {
	if(x<0)pc('-'),x=-x;
	int c[24]= {0};
	if(!x)return pc('0'),void();
	while(x)c[++c[0]]=x%10,x/=10;
	while(c[0])pc(c[c[0]--]+'0');
}
int fg=0;
int n,m,mt,f[5005][5005],to[5005],L[5005],R[5005];
int C(int x){return x*(x-1)/2;}
void add(int l,int r,int L1,int R1){
	if(l!=L1)fg=1;
	L[l-1]=L1-l+1,R[l-1]=R1-l,to[l-1]=r;
}
vector<pair<int,int>> vec;
void solve(){
	int lef=0,lim=0,L=0,R=0;
	for(auto p:vec){
		int l=p.first,r=p.second;
		if(l>lim){
			if(lef!=0)add(lef,lim,L,R);
			lef=l,lim=r,L=l,R=r;
		}else L=max(L,l),R=min(R,r),lim=max(lim,r);
	}
	if(lef!=0)add(lef,lim,L,R);
}
int main() {
	memset(f,-0x3f,sizeof(f));
	int Ti=getint();
	int tt=0;
	while(Ti--){
		n=getint(),m=getint(),mt=0,vec.clear();
		for(int i=1;i<=m;i++){
			int l=getint(),r=getint();
			vec.push_back(make_pair(l,r));
		}
		sort(vec.begin(),vec.end());
		solve();
		f[0][0]=0;
		for(int i=0;i<n;i++){
			if(L[i]){
				int len=to[i]-i,p=to[i];
				for(int j=0;j<=i;j++)
					for(int k=L[i];k<=R[i];k++)
						f[p][j+k]=max(f[p][j+k],f[i][j]+k*j+(len-k)*i+C(k)+C(len-k));
				i=p-1;
			}else{
				for(int j=0;j<=i;j++)
					f[i+1][j+1]=max(f[i+1][j+1],f[i][j]+j),
					f[i+1][j]=max(f[i+1][j],f[i][j]+i);
			}
		}
		int ans=-1;
		for(int i=1;i<=n;i++)
			for(int j=0;j<=i;j++)
				ans=max(ans,f[i][j]),f[i][j]=-0x3f3f3f3f;
		wrt(ans),pc('\n');
		for(int i=0;i<=n;i++)
			L[i]=R[i]=to[i]=0;
		if(++tt==47&&fg){
			cout<<n<<"_"<<m<<"_";
			for(auto p:vec){
				cout<<p.first<<"_"<<p.second<<"_";
			}
			exit(0);
		}
	}
	fwrite(wb,1,p2,stdout);
	return Loli Kon;
}

**Instructions:**
- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.
- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.
- Output the test input directly, not code to generate it.

**Output format:**
```plaintext
{test input}
```

Think step by step.

Response:
```plaintext
3
2 1
1 2
5 2
1 3
3 5
7 3
1 4
4 7
6 7
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Turtle and Inversions (Hard Version)
time_limit_ms: 2000
memory_limit_mb: 512
Description: This is a hard version of this problem. The differences between the versions are the constraint on $m$ and $r_i < l_{i + 1}$ holds for each $i$ from $1$ to $m - 1$ in the easy version. You can make hacks only if both versions of the problem are solved.

Turtle gives you $m$ intervals $[l_1, r_1], [l_2, r_2], \ldots, [l_m, r_m]$. He thinks that a permutation $p$ is interesting if there exists an integer $k_i$ for every interval ($l_i \le k_i < r_i$), and if he lets $a_i = \max\limits_{j = l_i}^{k_i} p_j, b_i = \min\limits_{j = k_i + 1}^{r_i} p_j$ for every integer $i$ from $1$ to $m$, the following condition holds:

$$\max\limits_{i = 1}^m a_i < \min\limits_{i = 1}^m b_i$$

Turtle wants you to calculate the maximum number of inversions of all interesting permutations of length $n$, or tell him if there is no interesting permutation.

An inversion of a permutation $p$ is a pair of integers $(i, j)$ ($1 \le i < j \le n$) such that $p_i > p_j$.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^3$). The description of the test cases follows.

The first line of each test case contains two integers $n, m$ ($2 \le n \le 5 \cdot 10^3, 0 \le m \le 5 \cdot 10^3$) — the length of the permutation and the number of intervals.

The $i$-th of the following $m$ lines contains two integers $l_i, r_i$ ($1 \le l_i < r_i \le n$) — the $i$-th interval. Note that there may exist identical intervals (i.e., there may exist two different indices $i, j$ such that $l_i = l_j$ and $r_i = r_j$).

It is guaranteed that the sum of $n$ over all test cases does not exceed $5 \cdot 10^3$ and the sum of $m$ over all test cases does not exceed $5 \cdot 10^3$.

Output: For each test case, if there is no interesting permutation, output a single integer $-1$.

Otherwise, output a single integer — the maximum number of inversions.

Examples:
input:
8
2 0
2 1
1 2
5 1
2 4
8 3
1 4
2 5
7 8
7 2
1 4
4 7
7 3
1 2
1 7
3 7
7 4
1 3
4 7
1 3
4 7
7 3
1 2
3 4
5 6
output:
1
0
8
18
-1
-1
15
15

Note: In the third test case, the interesting permutation with the maximum number of inversions is $[5, 2, 4, 3, 1]$.

In the fourth test case, the interesting permutation with the maximum number of inversions is $[4, 3, 8, 7, 6, 2, 1, 5]$. In this case, we can let $[k_1, k_2, k_3] = [2, 2, 7]$.

In the fifth and the sixth test case, it can be proven that there is no interesting permutation.

In the seventh test case, the interesting permutation with the maximum number of inversions is $[4, 7, 6, 3, 2, 1, 5]$. In this case, we can let $[k_1, k_2, k_3, k_4] = [1, 6, 1, 6]$.

In the eighth test case, the interesting permutation with the maximum number of inversions is $[4, 7, 3, 6, 2, 5, 1]$.

Buggy Code:
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<map>
#include<set>
#include<vector>
using namespace std;
const int inf=1e9,N=5005;
int T,K,n,m,f[N][N],val[N],isf[N];
struct node{int l,r;}a[N],st[N];
bool cmp(node A,node B){return A.l<B.l;}
signed main(){
	int i,j,c;
	scanf("%d",&T);
	while(T--){
		scanf("%d%d",&n,&m);
		int flg=0,tp=0;
		for(i=0;i<=n;i++){
			val[i]=0,isf[i]=0;
			for(j=0;j<=n;j++)f[i][j]=-inf;
		}
		for(i=1;i<=m;i++)scanf("%d%d",&a[i].l,&a[i].r);
		sort(a+1,a+1+m,cmp);
		if(m){
			node las=a[1];
			for(i=2;i<=m;i++){
				if(val[a[i].l]==1){flg=1;break;}
				int r=a[i].r;
				for(j=a[i].l+1;j<=a[i].r;j++){
					if(val[j]==1){r=j;break;}
				}
				a[i].r=r;
				if(las.r<a[i].l)st[++tp]=las,las=a[i];
				else{
					for(j=las.l;j<a[i].l;j++){
						if(!val[j])val[j]=-1;
						else if(val[j]!=-1){flg=1;break;}
					}
					for(j=las.r+1;j<=a[i].r;j++){
						if(!val[j])val[j]=1;
						else if(val[j]!=1){flg=1;break;}
					}
					if(flg)break;
					las=(node){a[i].l,min(las.r,a[i].r)};
				}
			}
			st[++tp]=las;
			if(flg){puts("-1");continue;}
		}
		f[0][0]=0;
		for(i=1;i<=tp;i++){
			isf[st[i].l]=st[i].r;
		//	printf("%d %d\n",st[i].l,st[i].r);
		}
		for(i=1;i<=n;i++){
			if(val[i]==-1){
				for(j=1;j<=i;j++){
					f[i][j]=max(f[i][j],f[i-1][j-1]+i-j);
				}
			}
			else if(val[i]==1){
				for(j=0;j<i;j++){
					f[i][j]=max(f[i][j],f[i-1][j]);
				}
			}
			else{
				if(!isf[i]){
					for(j=1;j<=i;j++)f[i][j]=max(f[i][j],f[i-1][j-1]+i-j);
					for(j=0;j<i;j++)f[i][j]=max(f[i][j],f[i-1][j]);
				}
				else{
					int len=isf[i]-i+1;
					if(len<=1){flg=1;break;}
					for(c=1;c<len;c++){//区间内小数个数
						for(j=c;j<=c+i-1;j++){
							int pnum=j-c;
							f[isf[i]][j]=max(f[isf[i]][j],f[i-1][pnum]+(i-1-pnum)*c);
						}
					}
					i=isf[i];
				}
			}
		}
		if(flg){puts("-1");continue;}
		int ans=0;
		for(i=0;i<=n;i++){
			ans=max(ans,f[n][i]+i*(i-1)/2+(n-i)*(n-i-1)/2);
		}
		printf("%d\n",ans);
	}

}

**Instructions:**
- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.
- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.
- Output the test input directly, not code to generate it.

**Output format:**
```plaintext
{test input}
```

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the given code, we need to consider edge cases and scenarios that can cause the code to fail. Here are the steps to think through this:

1. **Edge Cases**:
   - Test cases with no intervals.
   - Test cases with a single interval.
   - Test cases where intervals overlap in complex ways.
   - Test cases where intervals are non-overlapping.

2. **Complex Overlapping Intervals**:
   - Intervals that overlap in such a way that the condition for interesting permutations is difficult to satisfy.
   - Intervals that have very tight constraints (e.g., small ranges).

3. **Non-overlapping Intervals**:
   - Test cases where intervals are non-overlapping but still need to satisfy the condition for interesting permutations.

4. **Large and Small n and m**:
   - Test cases with small n and m to ensure the code handles small inputs correctly.
   - Test cases with large n and m to ensure the code handles large inputs within the given constraints.

Given these considerations, let's create a small-scale test input that is complete and valid:

```plaintext
4
3 1
1 3
4 2
1 2
3 4
5 3
1 2
2 3
4 5
6 4
1 2
3 4
5 6
6 7
```

This test input includes:
- A case with a single interval.
- A case with overlapping intervals.
- A case with non-overlapping intervals.
- A case with intervals that are almost non-overlapping but have a slight overlap.

These cases should help in exposing any bugs in the given code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Turtle and Inversions (Hard Version)
time_limit_ms: 2000
memory_limit_mb: 512
Description: This is a hard version of this problem. The differences between the versions are the constraint on $m$ and $r_i < l_{i + 1}$ holds for each $i$ from $1$ to $m - 1$ in the easy version. You can make hacks only if both versions of the problem are solved.

Turtle gives you $m$ intervals $[l_1, r_1], [l_2, r_2], \ldots, [l_m, r_m]$. He thinks that a permutation $p$ is interesting if there exists an integer $k_i$ for every interval ($l_i \le k_i < r_i$), and if he lets $a_i = \max\limits_{j = l_i}^{k_i} p_j, b_i = \min\limits_{j = k_i + 1}^{r_i} p_j$ for every integer $i$ from $1$ to $m$, the following condition holds:

$$\max\limits_{i = 1}^m a_i < \min\limits_{i = 1}^m b_i$$

Turtle wants you to calculate the maximum number of inversions of all interesting permutations of length $n$, or tell him if there is no interesting permutation.

An inversion of a permutation $p$ is a pair of integers $(i, j)$ ($1 \le i < j \le n$) such that $p_i > p_j$.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^3$). The description of the test cases follows.

The first line of each test case contains two integers $n, m$ ($2 \le n \le 5 \cdot 10^3, 0 \le m \le 5 \cdot 10^3$) — the length of the permutation and the number of intervals.

The $i$-th of the following $m$ lines contains two integers $l_i, r_i$ ($1 \le l_i < r_i \le n$) — the $i$-th interval. Note that there may exist identical intervals (i.e., there may exist two different indices $i, j$ such that $l_i = l_j$ and $r_i = r_j$).

It is guaranteed that the sum of $n$ over all test cases does not exceed $5 \cdot 10^3$ and the sum of $m$ over all test cases does not exceed $5 \cdot 10^3$.

Output: For each test case, if there is no interesting permutation, output a single integer $-1$.

Otherwise, output a single integer — the maximum number of inversions.

Examples:
input:
8
2 0
2 1
1 2
5 1
2 4
8 3
1 4
2 5
7 8
7 2
1 4
4 7
7 3
1 2
1 7
3 7
7 4
1 3
4 7
1 3
4 7
7 3
1 2
3 4
5 6
output:
1
0
8
18
-1
-1
15
15

Note: In the third test case, the interesting permutation with the maximum number of inversions is $[5, 2, 4, 3, 1]$.

In the fourth test case, the interesting permutation with the maximum number of inversions is $[4, 3, 8, 7, 6, 2, 1, 5]$. In this case, we can let $[k_1, k_2, k_3] = [2, 2, 7]$.

In the fifth and the sixth test case, it can be proven that there is no interesting permutation.

In the seventh test case, the interesting permutation with the maximum number of inversions is $[4, 7, 6, 3, 2, 1, 5]$. In this case, we can let $[k_1, k_2, k_3, k_4] = [1, 6, 1, 6]$.

In the eighth test case, the interesting permutation with the maximum number of inversions is $[4, 7, 3, 6, 2, 5, 1]$.

Buggy Code:
#include <bits/stdc++.h>

using namespace std;

int maxf(int a,int b){
    return a>b?a:b;
}

int minf(int a,int b){
    return a<b?a:b;
}

struct interval{
    int l;
    int r;
    int presekl;
    int presekr;
    void kopi(interval a){
        this->l=a.l;
        this->r=a.r;
        this->presekl=a.presekl;
        this->presekr=a.presekr;
    }
    bool spoji(interval a){//bool je ako uspe ili ne uspe
        //za interval a vazi a.l==a.presekl i a.r==a.presekr
        this->l=minf(this->l,a.l);
        this->r=maxf(this->r,a.r);
        this->presekl=maxf(this->presekl,a.l);
        this->presekr=minf(this->presekr,a.r);
        if(this->presekl>=this->presekr){
            return false;
        }
        return true;
    }
};

bool oksort(interval a,interval b){
    return a.l==b.l?a.r<=b.r:a.l<b.l;
}

void radi(){
    int n,m;
    cin>>n>>m;
    if(m==0){
        cout<<((n-1)*n/2)<<endl;
        return;
    }
    interval* a=new interval[m];
    int i;
    for(i=0;i<m;i++){
        cin>>a[i].l>>a[i].r;
        a[i].presekl=a[i].l;
        a[i].presekr=a[i].r;
    }
    sort(a,a+m,oksort);
    int j=1;
    i=1;
    while(i<m){
        if(a[i].l>a[j-1].r){
            a[j].kopi(a[i]);
            i++;
            j++;
        }
        else if(a[i].l==a[j-1].r){
            cout<<"-1"<<endl;
            delete[] a;
            return;
        }
        else{
            if(!(a[j-1].spoji(a[i]))){
                cout<<"-1"<<endl;
                delete[] a;
                return;
            }
            else{
                i++;
            }
        }
    }
    m=j;
//    for(i=0;i<m;i++){
//        cout<<a[i].l<<" "<<a[i].presekl<<" "<<a[i].presekr<<" "<<a[i].r<<endl;
//    }

    int* f=new int[n+1];
    int* fprethodno=new int[n+1];
    for(i=0;i<=n;i++){
        f[i]=-1;
        fprethodno[i]=-1;
    }
    fprethodno[0]=0;
    i=1;
    j=0;
    int t,k;
    while(i<=n){
        if(j<m && i==a[j].l){
            for(t=0;t<a[j].l;t++){//br1 pre ovoga
                if(fprethodno[t]!=-1){
                    //cout<<(a[j].presekl-a[j].l+1)<<" ";
                    for(k=a[j].presekl-a[j].l+1;k+a[j].l-1<a[j].presekr;k++){//br0
                        //cout<<k<<" ";
                        f[(t+a[j].r-a[j].l+1-k)]=maxf(f[(t+a[j].r-a[j].l+1-k)],fprethodno[t]+t*k);
                    }
                    //cout<<endl;
                }
            }
            i=a[j].r+1;
            j++;
        }
        else{
            for(t=0;t<i;t++){
                if(fprethodno[t]!=-1){
                    f[t]=maxf(f[t],fprethodno[t]+t);//0
                    f[t+1]=maxf(f[t+1],fprethodno[t]);//1
                }
            }
            i++;
        }
        for(t=0;t<=n;t++){
            //cout<<f[t]<<" ";
            fprethodno[t]=f[t];
            f[t]=-1;
        }
        //cout<<endl;
    }
    int maxx=-1;
    for(i=0;i<=n;i++){
        if(fprethodno[i]!=-1){
            maxx=maxf(maxx,fprethodno[i]+i*(i-1)/2+(n-i)*(n-i-1)/2);
        }
    }
    cout<<maxx<<endl;
    delete[] a;
    delete[] f;
    delete[] fprethodno;
}

int main()
{
    ios_base::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    int t;
    cin>>t;
    while(t>0){
        t--;
        radi();
    }
    //30
    return 0;
}

**Instructions:**
- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.
- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.
- Output the test input directly, not code to generate it.

**Output format:**
```plaintext
{test input}
```

Think step by step.

Response:
To create a challenging test input that exposes the bug in the given buggy code, we need to focus on edge cases and scenarios that can cause the algorithm to fail. Here are the steps to think through this:

1. **Identify Key Constraints**:
   - The intervals can overlap or be disjoint.
   - The intervals can be identical.
   - The intervals can be very small or very large.
   - The intervals can be such that the condition \( \max\limits_{i = 1}^m a_i < \min\limits_{i = 1}^m b_i \) is not possible.

2. **Consider Edge Cases**:
   - **Single Interval**: Test a single interval that covers the entire permutation length.
   - **Overlapping Intervals**: Test intervals that overlap in such a way that the condition for interesting permutations is difficult to satisfy.
   - **Disjoint Intervals**: Test intervals that are disjoint and far apart.
   - **Identical Intervals**: Test multiple identical intervals.
   - **Small Intervals**: Test very small intervals.
   - **Large Intervals**: Test very large intervals.
   - **No Intervals**: Test a case with no intervals.

3. **Construct the Test Input**:
   - **Single Interval**: This can help test the boundary conditions.
   - **Overlapping Intervals**: This can test the merging logic.
   - **Disjoint Intervals**: This can test the logic for handling disjoint intervals.
   - **Identical Intervals**: This can test the logic for handling multiple identical intervals.
   - **Small Intervals**: This can test the logic for handling small intervals.
   - **Large Intervals**: This can test the logic for handling large intervals.
   - **No Intervals**: This can test the base case where no intervals are provided.

Given these considerations, a small-scale test input that can expose the bug in the buggy code is:

```plaintext
4
5 1
1 5
5 2
1 3
3 5
5 3
1 2
3 4
5 5
5 0
```

### Explanation:
1. **Test Case 1 (Single Interval)**:
   - \( n = 5 \), \( m = 1 \)
   - Interval: \([1, 5]\)
   - This tests the case where a single interval covers the entire permutation length.

2. **Test Case 2 (Overlapping Intervals)**:
   - \( n = 5 \), \( m = 2 \)
   - Intervals: \([1, 3]\), \([3, 5]\)
   - This tests the case where intervals overlap.

3. **Test Case 3 (Disjoint Intervals)**:
   - \( n = 5 \), \( m = 3 \)
   - Intervals: \([1, 2]\), \([3, 4]\), \([5, 5]\)
   - This tests the case where intervals are disjoint and far apart.

4. **Test Case 4 (No Intervals)**:
   - \( n = 5 \), \( m = 0 \)
   - This tests the base case where no intervals are provided.

These test cases cover a range of scenarios that can help expose potential bugs in the algorithm.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Turtle and Inversions (Hard Version)
time_limit_ms: 2000
memory_limit_mb: 512
Description: This is a hard version of this problem. The differences between the versions are the constraint on $m$ and $r_i < l_{i + 1}$ holds for each $i$ from $1$ to $m - 1$ in the easy version. You can make hacks only if both versions of the problem are solved.

Turtle gives you $m$ intervals $[l_1, r_1], [l_2, r_2], \ldots, [l_m, r_m]$. He thinks that a permutation $p$ is interesting if there exists an integer $k_i$ for every interval ($l_i \le k_i < r_i$), and if he lets $a_i = \max\limits_{j = l_i}^{k_i} p_j, b_i = \min\limits_{j = k_i + 1}^{r_i} p_j$ for every integer $i$ from $1$ to $m$, the following condition holds:

$$\max\limits_{i = 1}^m a_i < \min\limits_{i = 1}^m b_i$$

Turtle wants you to calculate the maximum number of inversions of all interesting permutations of length $n$, or tell him if there is no interesting permutation.

An inversion of a permutation $p$ is a pair of integers $(i, j)$ ($1 \le i < j \le n$) such that $p_i > p_j$.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^3$). The description of the test cases follows.

The first line of each test case contains two integers $n, m$ ($2 \le n \le 5 \cdot 10^3, 0 \le m \le 5 \cdot 10^3$) — the length of the permutation and the number of intervals.

The $i$-th of the following $m$ lines contains two integers $l_i, r_i$ ($1 \le l_i < r_i \le n$) — the $i$-th interval. Note that there may exist identical intervals (i.e., there may exist two different indices $i, j$ such that $l_i = l_j$ and $r_i = r_j$).

It is guaranteed that the sum of $n$ over all test cases does not exceed $5 \cdot 10^3$ and the sum of $m$ over all test cases does not exceed $5 \cdot 10^3$.

Output: For each test case, if there is no interesting permutation, output a single integer $-1$.

Otherwise, output a single integer — the maximum number of inversions.

Examples:
input:
8
2 0
2 1
1 2
5 1
2 4
8 3
1 4
2 5
7 8
7 2
1 4
4 7
7 3
1 2
1 7
3 7
7 4
1 3
4 7
1 3
4 7
7 3
1 2
3 4
5 6
output:
1
0
8
18
-1
-1
15
15

Note: In the third test case, the interesting permutation with the maximum number of inversions is $[5, 2, 4, 3, 1]$.

In the fourth test case, the interesting permutation with the maximum number of inversions is $[4, 3, 8, 7, 6, 2, 1, 5]$. In this case, we can let $[k_1, k_2, k_3] = [2, 2, 7]$.

In the fifth and the sixth test case, it can be proven that there is no interesting permutation.

In the seventh test case, the interesting permutation with the maximum number of inversions is $[4, 7, 6, 3, 2, 1, 5]$. In this case, we can let $[k_1, k_2, k_3, k_4] = [1, 6, 1, 6]$.

In the eighth test case, the interesting permutation with the maximum number of inversions is $[4, 7, 3, 6, 2, 5, 1]$.

Buggy Code:
/**
 * (c) 2024 Diego Sogari
 */
#include <bits/stdc++.h>

using namespace std;

#ifdef ONLINE_JUDGE
#define debug
#else
#include "debug.h"
init();
#endif

template <typename T> ostream &operator<<(ostream &os, const vector<T> &a) {
  return ranges::for_each(a, [&os](auto &ai) { os << ai << ','; }), os;
}
void println1(auto &&...args) { ((cout << args << ';'), ...) << endl; }
void println(auto &&...args) { ((cout << args << ' '), ...) << endl; }

template <typename T> struct Num {
  T x;
  Num() { cin >> x; }
  Num(T a) : x(a) {}
  operator T &() { return x; }
  operator T() const { return x; }
};
using Int = Num<int>;

struct Digraph : vector<vector<int>> {
  const int n, m;
  Digraph(int n, int m = 0) : vector<vector<int>>(n + 1), n(n), m(m) {
    for (auto &[u, v] : vector<array<Int, 2>>(m)) {
      add(u, v);
    }
  }
  void add(int u, int v) { (*this)[u].push_back(v); }
};

void solve(int t) { // O(n + m)
  Int n, m;
  Digraph g(n, m);
  if (t == 180 && n != 6 && n != 10 && n != 13) {
    println1(n, m, g);
    return;
  }
  vector<int> a(n);
  for (int i = 1, l = 0, r = 0; i <= n; i++) // O(n + m)
  {
    int mn = n, mx = 0;
    for (auto &j : g[i]) {
      a[j - 1] = -1;
      mn = min(mn, j);
      mx = max(mx, j);
    }
    if (mx > 0) {
      if (i <= r) {
        for (int j = l + 1; j <= i; j++) {
          if (a[j - 1] < 0) {
            println(-1);
            return;
          }
          a[j - 1] = 1;
        }
        for (int j = min(mn, r) + 1; j <= max(mx, r); j++) {
          a[j - 1] = -1;
        }
      } else {
        a[i - 1] = 1;
      }
      l = i, r = max(r, mx);
    }
  }
  int ans = n * (n - 1) / 2, zeros = 0, ones = 0;
  for (int l = 0, r = n - 1; l <= r;) { // O(n)
    if (zeros < ones) {
      if (a[l] > 0) {
        zeros++;
      } else {
        ans -= zeros;
      }
      l++;
    } else {
      if (a[r] < 0) {
        ones++;
      } else {
        ans -= ones;
      }
      r--;
    }
  }
  println(ans - zeros * ones);
}

int main() {
  cin.tie(nullptr)->tie(nullptr)->sync_with_stdio(false);
  Int t;
  for (int i = 1; i <= t; i++) {
    solve(i);
  }
}

**Instructions:**
- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.
- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.
- Output the test input directly, not code to generate it.

**Output format:**
```plaintext
{test input}
```

Think step by step.

Response:
To expose the bug in the provided code, we need to focus on edge cases or scenarios that might cause the code to fail. The code appears to handle the construction of a permutation and the calculation of the maximum number of inversions. The main issues could arise from how it processes intervals and the conditions for forming an interesting permutation.

### Step-by-Step Analysis:

1. **Interval Overlap and Disjoint Intervals:**
   - The code should handle intervals that overlap and intervals that are disjoint.
   - If there are overlapping intervals, the code should ensure that the conditions for forming an interesting permutation are met.

2. **Empty Intervals:**
   - The code should handle cases where there are no intervals (i.e., `m = 0`).

3. **Single Interval:**
   - The code should handle cases where there is only one interval.

4. **Multiple Intervals:**
   - The code should handle cases where there are multiple intervals, and the intervals may or may not overlap.

5. **Maximum and Minimum Values:**
   - The code should correctly handle the maximum and minimum values within the intervals.

### Test Input Generation:

Let's generate a small-scale test input that covers these scenarios:

1. **Test Case 1: No Intervals (m = 0)**
   - This should return the maximum number of inversions for a permutation of length `n`.

2. **Test Case 2: Single Interval**
   - This should check if the code can handle a single interval and form an interesting permutation.

3. **Test Case 3: Overlapping Intervals**
   - This should check if the code can handle overlapping intervals and form an interesting permutation.

4. **Test Case 4: Disjoint Intervals**
   - This should check if the code can handle disjoint intervals and form an interesting permutation.

### Test Input:

```plaintext
4
5 0
5 1
1 3
5 2
1 3
3 5
5 2
1 2
4 5
```

### Explanation:

1. **Test Case 1:**
   - `n = 5`, `m = 0`
   - There are no intervals, so the maximum number of inversions is `10` (for the permutation `[5, 4, 3, 2, 1]`).

2. **Test Case 2:**
   - `n = 5`, `m = 1`
   - Interval: `[1, 3]`
   - The interesting permutation with the maximum number of inversions is `[5, 4, 3, 2, 1]` (8 inversions).

3. **Test Case 3:**
   - `n = 5`, `m = 2`
   - Intervals: `[1, 3]`, `[3, 5]`
   - The interesting permutation with the maximum number of inversions is `[5, 4, 3, 2, 1]` (8 inversions).

4. **Test Case 4:**
   - `n = 5`, `m = 2`
   - Intervals: `[1, 2]`, `[4, 5]`
   - The interesting permutation with the maximum number of inversions is `[5, 4, 3, 2, 1]` (8 inversions).

This test input should help expose any issues in the code, particularly in how it handles different types of intervals and the construction of the permutation.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Turtle and Inversions (Hard Version)
time_limit_ms: 2000
memory_limit_mb: 512
Description: This is a hard version of this problem. The differences between the versions are the constraint on $m$ and $r_i < l_{i + 1}$ holds for each $i$ from $1$ to $m - 1$ in the easy version. You can make hacks only if both versions of the problem are solved.

Turtle gives you $m$ intervals $[l_1, r_1], [l_2, r_2], \ldots, [l_m, r_m]$. He thinks that a permutation $p$ is interesting if there exists an integer $k_i$ for every interval ($l_i \le k_i < r_i$), and if he lets $a_i = \max\limits_{j = l_i}^{k_i} p_j, b_i = \min\limits_{j = k_i + 1}^{r_i} p_j$ for every integer $i$ from $1$ to $m$, the following condition holds:

$$\max\limits_{i = 1}^m a_i < \min\limits_{i = 1}^m b_i$$

Turtle wants you to calculate the maximum number of inversions of all interesting permutations of length $n$, or tell him if there is no interesting permutation.

An inversion of a permutation $p$ is a pair of integers $(i, j)$ ($1 \le i < j \le n$) such that $p_i > p_j$.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^3$). The description of the test cases follows.

The first line of each test case contains two integers $n, m$ ($2 \le n \le 5 \cdot 10^3, 0 \le m \le 5 \cdot 10^3$) — the length of the permutation and the number of intervals.

The $i$-th of the following $m$ lines contains two integers $l_i, r_i$ ($1 \le l_i < r_i \le n$) — the $i$-th interval. Note that there may exist identical intervals (i.e., there may exist two different indices $i, j$ such that $l_i = l_j$ and $r_i = r_j$).

It is guaranteed that the sum of $n$ over all test cases does not exceed $5 \cdot 10^3$ and the sum of $m$ over all test cases does not exceed $5 \cdot 10^3$.

Output: For each test case, if there is no interesting permutation, output a single integer $-1$.

Otherwise, output a single integer — the maximum number of inversions.

Examples:
input:
8
2 0
2 1
1 2
5 1
2 4
8 3
1 4
2 5
7 8
7 2
1 4
4 7
7 3
1 2
1 7
3 7
7 4
1 3
4 7
1 3
4 7
7 3
1 2
3 4
5 6
output:
1
0
8
18
-1
-1
15
15

Note: In the third test case, the interesting permutation with the maximum number of inversions is $[5, 2, 4, 3, 1]$.

In the fourth test case, the interesting permutation with the maximum number of inversions is $[4, 3, 8, 7, 6, 2, 1, 5]$. In this case, we can let $[k_1, k_2, k_3] = [2, 2, 7]$.

In the fifth and the sixth test case, it can be proven that there is no interesting permutation.

In the seventh test case, the interesting permutation with the maximum number of inversions is $[4, 7, 6, 3, 2, 1, 5]$. In this case, we can let $[k_1, k_2, k_3, k_4] = [1, 6, 1, 6]$.

In the eighth test case, the interesting permutation with the maximum number of inversions is $[4, 7, 3, 6, 2, 5, 1]$.

Buggy Code:
#include<bits/stdc++.h>
using namespace std;

typedef long long ll;
const int N=5003,INF=1e9;
int n,m,pl[N],pr[N],a[N],kl[N],kr[N],f[N],g[N],pos[N];
int Val(int x){return x*(x-1)/2;}
int Sol()
{
	for(int i=0;i<=n;i++)f[i]=g[i]=-INF,pos[i]=0;
	f[0]=0;int ans=0;
	for(int i=1;i<=m;i++)
	    for(int j=kl[i];j<=kr[i];j++)pos[j]=i;
	for(int i=1;i<=n;i++)
	{
		memcpy(g,f,sizeof(g));memset(f,0xcf,sizeof(f));
		if(!pos[i])
		{
			for(int j=0;j<=i;j++)
			{
			    if(a[i]!=0)f[j+1]=max(f[j+1],g[j]);
				if(a[i]!=1)f[j]=max(f[j],g[j]+j);
			}
			continue;
		}
		int len=kr[pos[i]]-kl[pos[i]]+1;
		for(int j=0;j<=i;j++)for(int k=0;k<=len;k++)f[j+k]=max(f[j+k],g[j]+j*(len-k));
		i=kr[pos[i]];
	}
	for(int i=0;i<=n;i++)ans=max(ans,f[i]+Val(i)+Val(n-i));
	return ans;
}
void Solve()
{
	cin>>n>>m;
	for(int i=1;i<=n;i++)a[i]=-1;
	for(int i=1;i<=m;i++)
	{
		cin>>pl[i]>>pr[i];a[pl[i]]=0;a[pr[i]]=1;
		for(int j=pl[i]+1;j<pr[i];j++)a[j]=2;
	}
	for(int i=1;i<=m;i++)for(int j=1;j<=m;j++)if(i!=j&&pl[i]<=pl[j]&&pr[i]>=pl[j])
	{
		for(int k=pl[i];k<pl[j];k++)a[k]=0;
		for(int k=min(pr[i],pr[j])+1;k<=max(pr[i],pr[j]);k++)a[k]=1;
	}
	for(int i=1;i<=m;i++)
	{
		int t0=0,t1=0,len=pr[i]-pl[i]+1;
		for(int j=pl[i];j<=pr[i];j++)t0+=a[j]==0,t1+=a[j]==0;
		if(t1==len||t1==len){cout<<-1<<endl;return;}
	}
	m=0;
	for(int i=1;i<=n;i++)if(a[i]==2)
	{
		int j=i;
		while(j<n&&a[j+1]==2)j++;
		m++;kl[m]=i;kr[m]=j;i=j;
	}
	cout<<Sol()<<endl;
}
int main()
{
	int T;cin>>T;
	while(T--)Solve();
	return 0;
}

**Instructions:**
- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.
- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.
- Output the test input directly, not code to generate it.

**Output format:**
```plaintext
{test input}
```

Think step by step.

Response:
```plaintext
2
3 2
1 2
2 3
5 3
1 2
2 3
3 4
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Turtle and Inversions (Hard Version)
time_limit_ms: 2000
memory_limit_mb: 512
Description: This is a hard version of this problem. The differences between the versions are the constraint on $m$ and $r_i < l_{i + 1}$ holds for each $i$ from $1$ to $m - 1$ in the easy version. You can make hacks only if both versions of the problem are solved.

Turtle gives you $m$ intervals $[l_1, r_1], [l_2, r_2], \ldots, [l_m, r_m]$. He thinks that a permutation $p$ is interesting if there exists an integer $k_i$ for every interval ($l_i \le k_i < r_i$), and if he lets $a_i = \max\limits_{j = l_i}^{k_i} p_j, b_i = \min\limits_{j = k_i + 1}^{r_i} p_j$ for every integer $i$ from $1$ to $m$, the following condition holds:

$$\max\limits_{i = 1}^m a_i < \min\limits_{i = 1}^m b_i$$

Turtle wants you to calculate the maximum number of inversions of all interesting permutations of length $n$, or tell him if there is no interesting permutation.

An inversion of a permutation $p$ is a pair of integers $(i, j)$ ($1 \le i < j \le n$) such that $p_i > p_j$.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^3$). The description of the test cases follows.

The first line of each test case contains two integers $n, m$ ($2 \le n \le 5 \cdot 10^3, 0 \le m \le 5 \cdot 10^3$) — the length of the permutation and the number of intervals.

The $i$-th of the following $m$ lines contains two integers $l_i, r_i$ ($1 \le l_i < r_i \le n$) — the $i$-th interval. Note that there may exist identical intervals (i.e., there may exist two different indices $i, j$ such that $l_i = l_j$ and $r_i = r_j$).

It is guaranteed that the sum of $n$ over all test cases does not exceed $5 \cdot 10^3$ and the sum of $m$ over all test cases does not exceed $5 \cdot 10^3$.

Output: For each test case, if there is no interesting permutation, output a single integer $-1$.

Otherwise, output a single integer — the maximum number of inversions.

Examples:
input:
8
2 0
2 1
1 2
5 1
2 4
8 3
1 4
2 5
7 8
7 2
1 4
4 7
7 3
1 2
1 7
3 7
7 4
1 3
4 7
1 3
4 7
7 3
1 2
3 4
5 6
output:
1
0
8
18
-1
-1
15
15

Note: In the third test case, the interesting permutation with the maximum number of inversions is $[5, 2, 4, 3, 1]$.

In the fourth test case, the interesting permutation with the maximum number of inversions is $[4, 3, 8, 7, 6, 2, 1, 5]$. In this case, we can let $[k_1, k_2, k_3] = [2, 2, 7]$.

In the fifth and the sixth test case, it can be proven that there is no interesting permutation.

In the seventh test case, the interesting permutation with the maximum number of inversions is $[4, 7, 6, 3, 2, 1, 5]$. In this case, we can let $[k_1, k_2, k_3, k_4] = [1, 6, 1, 6]$.

In the eighth test case, the interesting permutation with the maximum number of inversions is $[4, 7, 3, 6, 2, 5, 1]$.

Buggy Code:
#include<bits/stdc++.h>
using namespace std;
#define task "a"
#define se second
#define fi first
#define ll long long
#define ii pair<ll, ll>
const long mxN = 5e3 + 7;

int l[mxN], r[mxN];
ll dd[mxN];

int main()
{
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);
    //freopen(task".INP", "r", stdin);
    //freopen(task".OUT", "w", stdout);

    int tst;
    cin >> tst;
    while (tst--)
    {
        int n, m;
        cin >> n >> m;
        for (int i = 1; i <= n; i++)
            dd[i] = 0;

        vector<int> val, en;
        val.push_back(0);
        for (int i = 1; i <= m; i++)
        {
            cin >> l[i] >> r[i];
            dd[r[i]] = 1;
            val.push_back(l[i]);
        }
        sort(val.begin(), val.end());
        val.erase(unique(val.begin(), val.end()), val.end());

        for (int i = 1; i <= m; i++)
        {
            bool tt = 0;
            for (int j = l[i]; j <= r[i]; j++)
            {
                tt |= dd[j];
                dd[j] = tt;
            }
        }

        ll ans = 0;
        for (int i : val)
        {
            if (dd[i])
                ans = -1;
        }

        if (ans == -1)
        {
            cout << -1 << '\n';
            continue;
        }

        for (int i = val.size() - 1; i <= n - val.size() + 1; i++)
        {
            ll res = 0;
            int cur = val.size() - 1, rem = i;
            for (int j = n; j >= 1; j--)
            {
                if (dd[j])
                    continue;

                if (j == val[cur])
                    cur--;

                if (cur < rem)
                {
                    rem--;
                    res += j - 1;
                }
            }
            ans = max(res + (n - i - 1) * (n - i) / 2, ans);
        }
        cout << ans << '\n';
    }
}

**Instructions:**
- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.
- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.
- Output the test input directly, not code to generate it.

**Output format:**
```plaintext
{test input}
```

Think step by step.

Response:
```plaintext
3
2 0
2 1
1 2
5 1
2 4
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Turtle and Inversions (Hard Version)
time_limit_ms: 2000
memory_limit_mb: 512
Description: This is a hard version of this problem. The differences between the versions are the constraint on $m$ and $r_i < l_{i + 1}$ holds for each $i$ from $1$ to $m - 1$ in the easy version. You can make hacks only if both versions of the problem are solved.

Turtle gives you $m$ intervals $[l_1, r_1], [l_2, r_2], \ldots, [l_m, r_m]$. He thinks that a permutation $p$ is interesting if there exists an integer $k_i$ for every interval ($l_i \le k_i < r_i$), and if he lets $a_i = \max\limits_{j = l_i}^{k_i} p_j, b_i = \min\limits_{j = k_i + 1}^{r_i} p_j$ for every integer $i$ from $1$ to $m$, the following condition holds:

$$\max\limits_{i = 1}^m a_i < \min\limits_{i = 1}^m b_i$$

Turtle wants you to calculate the maximum number of inversions of all interesting permutations of length $n$, or tell him if there is no interesting permutation.

An inversion of a permutation $p$ is a pair of integers $(i, j)$ ($1 \le i < j \le n$) such that $p_i > p_j$.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^3$). The description of the test cases follows.

The first line of each test case contains two integers $n, m$ ($2 \le n \le 5 \cdot 10^3, 0 \le m \le 5 \cdot 10^3$) — the length of the permutation and the number of intervals.

The $i$-th of the following $m$ lines contains two integers $l_i, r_i$ ($1 \le l_i < r_i \le n$) — the $i$-th interval. Note that there may exist identical intervals (i.e., there may exist two different indices $i, j$ such that $l_i = l_j$ and $r_i = r_j$).

It is guaranteed that the sum of $n$ over all test cases does not exceed $5 \cdot 10^3$ and the sum of $m$ over all test cases does not exceed $5 \cdot 10^3$.

Output: For each test case, if there is no interesting permutation, output a single integer $-1$.

Otherwise, output a single integer — the maximum number of inversions.

Examples:
input:
8
2 0
2 1
1 2
5 1
2 4
8 3
1 4
2 5
7 8
7 2
1 4
4 7
7 3
1 2
1 7
3 7
7 4
1 3
4 7
1 3
4 7
7 3
1 2
3 4
5 6
output:
1
0
8
18
-1
-1
15
15

Note: In the third test case, the interesting permutation with the maximum number of inversions is $[5, 2, 4, 3, 1]$.

In the fourth test case, the interesting permutation with the maximum number of inversions is $[4, 3, 8, 7, 6, 2, 1, 5]$. In this case, we can let $[k_1, k_2, k_3] = [2, 2, 7]$.

In the fifth and the sixth test case, it can be proven that there is no interesting permutation.

In the seventh test case, the interesting permutation with the maximum number of inversions is $[4, 7, 6, 3, 2, 1, 5]$. In this case, we can let $[k_1, k_2, k_3, k_4] = [1, 6, 1, 6]$.

In the eighth test case, the interesting permutation with the maximum number of inversions is $[4, 7, 3, 6, 2, 5, 1]$.

Buggy Code:
#include <bits/stdc++.h>
#define all(x) x.begin(), x.end()
#define pb push_back
#define eb emplace_back
#define rz resize
#define MP make_pair
#define MT make_tuple
#define IT iterator
#define fi first
#define se second
#define For(i, a, b) for (int i = (int)(a); i <= (int)(b); ++i)
#define Rep(i, a, b) for (int i = (int)(a); i >= (int)(b); --i)
#define CLR(a, v) memset(a, v, sizeof(a))
#define CPY(a, b) memcpy(a, b, sizeof(a))
#define debug cerr << "ztxakking\n"
#define y0 ztxaknoi
#define y1 ztxakioi
using namespace std;
using ll = long long;
using ld = long double;
using ull = unsigned long long;
using uint = unsigned int;
using pii = pair<int, int>;
using pll = pair<ll, ll>;
using pli = pair<ll, int>;
using pil = pair<int, ll>;
using vi = vector<int>;
template<typename T>
using V = vector<T>;
mt19937 rnd(chrono::steady_clock::now().time_since_epoch().count());
const int N = 5e3 + 7, Inf = 0x3f3f3f3f;
int n, m, f[N], g[N], ans;
pii a[N];
tuple<int, int, int> b[N];
bitset<N> e[N], vis;
vi A;
void dfs(int u) {
  if (vis[u]) return ;
  vis[u] = 1, A.pb(u);
  for (int i = e[u]._Find_first(); i <= n; i = e[u]._Find_next(i)) dfs(i);
}
void Main() {
  cin >> n >> m;
  For(i, 1, m) e[i].reset();
  For(i, 1, m) {
    auto &[l, r] = a[i];
    cin >> l >> r;
  }
  For(i, 1, m) For(j, i + 1, m) {
    auto [l1, r1] = a[i];
    auto [l2, r2] = a[j];
    if (max(l1, l2) <= min(r1, r2)) e[i][j] = e[j][i] = 1;
  }
  For(i, 1, m) vis[i] = 0;
  For(i, 1, n) b[i] = {0, 0, 0};
  For(i, 1, m) if (!vis[i]) {
    A.clear(), dfs(i);
    For(x, 0, A.size() - 1) For(y, x + 1, A.size() - 1) {
      auto [l1, r1] = a[A[x]];
      auto [l2, r2] = a[A[y]];
      if (max(l1, l2) >= min(r1, r2)) return cout << "-1\n", void();
    }
    int lmi = n, lmx = 0, rmi = n, rmx = 0;
    for (int x : A) {
      auto [l, r] = a[x];
      lmi = min(lmi, l), lmx = max(lmx, l);
      rmi = min(rmi, r), rmx = max(rmx, r);
    }
    b[lmi] = {lmx, rmi, rmx};
  }
  For(i, 1, n) vis[i] = 0;
  For(i, 1, m) {
    auto [l, r] = a[i];
    For(j, l, r) vis[j] = 1;
  }
  For(i, 0, n) f[i] = g[i] = -Inf;
  f[0] = 0;
  For(i, 1, n) {
    For(j, 0, i - 1) g[j] = -Inf;
    if (!vis[i]) {
      For(j, 0, i - 1) {
        g[j] = max(g[j], f[j] + j);
        g[j + 1] = max(g[j + 1], f[j]);
      }
    } else {
      auto [l, r1, r2] = b[i];
      For(j, 0, i - 1) {
        For(s, l + 1, r1) {
          int x = s - i, y = r2 - s + 1;
          g[j + y] = max(g[j + y], f[j] + x * j);
        }
      }
      i = r2;
    }
    For(j, 0, i) f[j] = g[j];
  }
  ans = -Inf;
  For(i, 0, n) {
    ans = max(ans, f[i] + i * (i - 1) / 2 + (n - i) * (n - i - 1) / 2);
  }
  cout << ans << '\n';
}
int main() {
  ios::sync_with_stdio(0), cin.tie(0);
  int t; cin >> t;
  while (t--) Main();
  cerr << (double)(clock()) / CLOCKS_PER_SEC << '\n';
  return 0;
}

**Instructions:**
- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.
- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.
- Output the test input directly, not code to generate it.

**Output format:**
```plaintext
{test input}
```

Think step by step.

Response:
```plaintext
1
5 2
1 3
3 5
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Turtle and Inversions (Hard Version)
time_limit_ms: 2000
memory_limit_mb: 512
Description: This is a hard version of this problem. The differences between the versions are the constraint on $m$ and $r_i < l_{i + 1}$ holds for each $i$ from $1$ to $m - 1$ in the easy version. You can make hacks only if both versions of the problem are solved.

Turtle gives you $m$ intervals $[l_1, r_1], [l_2, r_2], \ldots, [l_m, r_m]$. He thinks that a permutation $p$ is interesting if there exists an integer $k_i$ for every interval ($l_i \le k_i < r_i$), and if he lets $a_i = \max\limits_{j = l_i}^{k_i} p_j, b_i = \min\limits_{j = k_i + 1}^{r_i} p_j$ for every integer $i$ from $1$ to $m$, the following condition holds:

$$\max\limits_{i = 1}^m a_i < \min\limits_{i = 1}^m b_i$$

Turtle wants you to calculate the maximum number of inversions of all interesting permutations of length $n$, or tell him if there is no interesting permutation.

An inversion of a permutation $p$ is a pair of integers $(i, j)$ ($1 \le i < j \le n$) such that $p_i > p_j$.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^3$). The description of the test cases follows.

The first line of each test case contains two integers $n, m$ ($2 \le n \le 5 \cdot 10^3, 0 \le m \le 5 \cdot 10^3$) — the length of the permutation and the number of intervals.

The $i$-th of the following $m$ lines contains two integers $l_i, r_i$ ($1 \le l_i < r_i \le n$) — the $i$-th interval. Note that there may exist identical intervals (i.e., there may exist two different indices $i, j$ such that $l_i = l_j$ and $r_i = r_j$).

It is guaranteed that the sum of $n$ over all test cases does not exceed $5 \cdot 10^3$ and the sum of $m$ over all test cases does not exceed $5 \cdot 10^3$.

Output: For each test case, if there is no interesting permutation, output a single integer $-1$.

Otherwise, output a single integer — the maximum number of inversions.

Examples:
input:
8
2 0
2 1
1 2
5 1
2 4
8 3
1 4
2 5
7 8
7 2
1 4
4 7
7 3
1 2
1 7
3 7
7 4
1 3
4 7
1 3
4 7
7 3
1 2
3 4
5 6
output:
1
0
8
18
-1
-1
15
15

Note: In the third test case, the interesting permutation with the maximum number of inversions is $[5, 2, 4, 3, 1]$.

In the fourth test case, the interesting permutation with the maximum number of inversions is $[4, 3, 8, 7, 6, 2, 1, 5]$. In this case, we can let $[k_1, k_2, k_3] = [2, 2, 7]$.

In the fifth and the sixth test case, it can be proven that there is no interesting permutation.

In the seventh test case, the interesting permutation with the maximum number of inversions is $[4, 7, 6, 3, 2, 1, 5]$. In this case, we can let $[k_1, k_2, k_3, k_4] = [1, 6, 1, 6]$.

In the eighth test case, the interesting permutation with the maximum number of inversions is $[4, 7, 3, 6, 2, 5, 1]$.

Buggy Code:
#include <bits/stdc++.h>
#define ll long long
#define fi first
#define se second
#define mpr make_pair
#define ull unsigned ll
#define pb push_back
using namespace std;

int timer;
void solve() {
    ++timer;
    int n, m;
    cin >> n >> m;
    vector to(n + 1, 0);
    vector rg(n + 1, mpr(0, 0));
    vector<pair<int, int> > g;
    while (m--) {
        int a, b;
        cin >> a >> b;
        g.pb(mpr(a, b));
    }
    sort(g.begin(), g.end());
    for (int i = 0; i < g.size(); i++) {
        int j = i + 1;
        int mx = g[i].se;
        while (j < g.size() && g[j].fi <= mx) {
            mx = max(g[j].se, mx);
            j++;
        }
        int L = g[i].fi;
        int R = mx;
        int l = L, r = R;
        for (int k = i; k < j; k++) {
            l = max(l, g[k].fi);
            r = min(r, g[k].se);
        }
        to[L] = R;
        rg[L].fi = l - L + 1;
        rg[L].se = r - L;
        // cout << rg[L].fi << " " << rg[L].se << "  shit\n";
        i = j - 1;
    }
    auto ck = [&](int& a, int b) {
        a = max(a, b);
    };
    vector f(n + 1, vector(n + 1, -(int)1e9));
    f[0][0] = 0;
    for (int i = 1; i <= n; i++) {
        int old = i - 1;
        if (to[i] == 0) {
            for (int j = 0; j <= i; j++) {
                ck(f[i][j], f[old][j] + j);
                if (j - 1 >= 0) {
                    ck(f[i][j], f[old][j - 1]);
                }
            }
        } else {
            int len = to[i] - i + 1;
            auto [L, R] = rg[i];
            i = to[i];
            for (int j = 0; j <= i; j++) {
                for (int a = L; a <= R; a++) {
                    if (j - a >= 0) {
                        ck(f[i][j], f[old][j - a] + (len - a) * (j - a));
                        // cout<<i<<" "<<j<<" "<<old<<" "<<j-a<<" "<<f[i][j]<<"  shit\n";
                    }
                }
            }
        }
    }
    int ans = -1e9;
    for (int i = 0; i <= n; i++) {
        // cout<<n<<" "<<i<<" "<<f[n][i]<<" "<<i*(i-1)/2<<" "<<(n-i)*(n-i-1)/2<<"\n";
        ans = max(ans, f[n][i] + i * (i - 1) / 2 + (n - i) * (n - i - 1) / 2);
    }
    if (timer == 47 && ans == 36) {
        cout << "!\n";
        cout << n << " " << m << "\n";
        for (int i = 1; i <= g.size(); i++) {
            cout << g[i - 1].fi << " " << g[i - 1].se << "\n";
        }
        exit(0);
    }
    cout << max(ans, -1) << "\n";
}
int main() {
    ios::sync_with_stdio(0);
    cin.tie(0), cout.tie(0);
    int t = 1;
    cin >> t;
    while (t--) solve();
    return 0;
}
/*
1
7 4 1
2 3 6 5 8 9 8
2 7
*/

**Instructions:**
- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.
- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.
- Output the test input directly, not code to generate it.

**Output format:**
```plaintext
{test input}
```

Think step by step.

Response:
```plaintext
1
5 2
1 3
3 5
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Turtle and Inversions (Hard Version)
time_limit_ms: 2000
memory_limit_mb: 512
Description: This is a hard version of this problem. The differences between the versions are the constraint on $m$ and $r_i < l_{i + 1}$ holds for each $i$ from $1$ to $m - 1$ in the easy version. You can make hacks only if both versions of the problem are solved.

Turtle gives you $m$ intervals $[l_1, r_1], [l_2, r_2], \ldots, [l_m, r_m]$. He thinks that a permutation $p$ is interesting if there exists an integer $k_i$ for every interval ($l_i \le k_i < r_i$), and if he lets $a_i = \max\limits_{j = l_i}^{k_i} p_j, b_i = \min\limits_{j = k_i + 1}^{r_i} p_j$ for every integer $i$ from $1$ to $m$, the following condition holds:

$$\max\limits_{i = 1}^m a_i < \min\limits_{i = 1}^m b_i$$

Turtle wants you to calculate the maximum number of inversions of all interesting permutations of length $n$, or tell him if there is no interesting permutation.

An inversion of a permutation $p$ is a pair of integers $(i, j)$ ($1 \le i < j \le n$) such that $p_i > p_j$.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^3$). The description of the test cases follows.

The first line of each test case contains two integers $n, m$ ($2 \le n \le 5 \cdot 10^3, 0 \le m \le 5 \cdot 10^3$) — the length of the permutation and the number of intervals.

The $i$-th of the following $m$ lines contains two integers $l_i, r_i$ ($1 \le l_i < r_i \le n$) — the $i$-th interval. Note that there may exist identical intervals (i.e., there may exist two different indices $i, j$ such that $l_i = l_j$ and $r_i = r_j$).

It is guaranteed that the sum of $n$ over all test cases does not exceed $5 \cdot 10^3$ and the sum of $m$ over all test cases does not exceed $5 \cdot 10^3$.

Output: For each test case, if there is no interesting permutation, output a single integer $-1$.

Otherwise, output a single integer — the maximum number of inversions.

Examples:
input:
8
2 0
2 1
1 2
5 1
2 4
8 3
1 4
2 5
7 8
7 2
1 4
4 7
7 3
1 2
1 7
3 7
7 4
1 3
4 7
1 3
4 7
7 3
1 2
3 4
5 6
output:
1
0
8
18
-1
-1
15
15

Note: In the third test case, the interesting permutation with the maximum number of inversions is $[5, 2, 4, 3, 1]$.

In the fourth test case, the interesting permutation with the maximum number of inversions is $[4, 3, 8, 7, 6, 2, 1, 5]$. In this case, we can let $[k_1, k_2, k_3] = [2, 2, 7]$.

In the fifth and the sixth test case, it can be proven that there is no interesting permutation.

In the seventh test case, the interesting permutation with the maximum number of inversions is $[4, 7, 6, 3, 2, 1, 5]$. In this case, we can let $[k_1, k_2, k_3, k_4] = [1, 6, 1, 6]$.

In the eighth test case, the interesting permutation with the maximum number of inversions is $[4, 7, 3, 6, 2, 5, 1]$.

Buggy Code:
// They say that life is always easier
// After you let yourself come undone
#include <bits/stdc++.h>
using namespace std;
const int N = 5e3 + 10;
const int INF = 1e9;
struct Node {
	int l, r, lef, rig;
} a[N], b[N];
int n, m, top, rig[N], L[N], R[N];
int f[N], g[N];
void Solve() {
	cin >> n >> m;
	for(int i = 1; i <= m; ++i)
		cin >> b[i].l >> b[i].r, b[i].lef = b[i].l, b[i].rig = b[i].r;
	sort(b + 1, b + m + 1, [](Node a, Node b) {
		if(a.l != b.l) return a.l < b.l;
		return a.r < b.r;
	});
	if(m) a[top = 1] = b[1];
	else top = 0;
	for(int i = 1; i <= m; ++i)
		if(a[top].r < b[i].l) a[++top] = b[i];
		else if(a[top].r < b[i].r)
			a[top].r = b[i].r, a[top].lef = max(a[top].lef, b[i].l);
		else {
			a[top].lef = max(a[top].lef, b[i].l);
			a[top].rig = min(a[top].rig, b[i].r);
		}
	bool isOK = 1;
	for(int i = 1; i <= top; ++i) {
		if(a[i].lef >= a[i].rig) isOK = 0;
		rig[a[i].l] = a[i].r, L[a[i].l] = a[i].lef, R[a[i].l] = a[i].rig;
	}
	if(!isOK) return printf("-1\n"), void();
	for(int i = 0; i <= n; ++i) f[i] = -INF;
	f[0] = 0;
	for(int i = 1; i <= n; ++i) {
		for(int j = 0; j <= n; ++j) g[j] = -INF;
		if(rig[i]) {
			int len = rig[i] - i + 1;
			for(int j = 0; j <= n; ++j) if(f[j] != -INF)
				for(int k = L[i]; k < R[i]; ++k) {
					int x = k - i + 1, y = rig[i] - i + 1 - x;
					g[j + y] = max(g[j + y], f[j] + j * x);
				}
			i = rig[i];
		}
		else {
			for(int j = 0; j <= n; ++j) if(f[j] != -INF) {
				g[j] = max(g[j], f[j] + j);
				g[j + 1] = max(g[j + 1], f[j]);
			}
		}
		for(int j = 0; j <= n; ++j) f[j] = g[j];
	}
	int ans = -INF;
	for(int i = 1; i <= n; ++i) {
		int j = n - i;
		ans = max(ans, f[i] + i * (i - 1) / 2 + j * (j - 1) / 2);
	}
	printf("%d\n", ans);
	for(int i = 1; i <= n; ++i) rig[i] = 0;
}
int main() {
	//freopen("data.in", "r", stdin);
	//freopen("code.out", "w", stdout);
	cin.tie(0)->sync_with_stdio(0);
	int t; cin >> t;
	while(t--) Solve();
	return 0;
}

**Instructions:**
- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.
- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.
- Output the test input directly, not code to generate it.

**Output format:**
```plaintext
{test input}
```

Think step by step.

Response:
```plaintext
3
2 1
1 2
3 1
1 3
5 2
1 2
4 5
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Turtle and Inversions (Hard Version)
time_limit_ms: 2000
memory_limit_mb: 512
Description: This is a hard version of this problem. The differences between the versions are the constraint on $m$ and $r_i < l_{i + 1}$ holds for each $i$ from $1$ to $m - 1$ in the easy version. You can make hacks only if both versions of the problem are solved.

Turtle gives you $m$ intervals $[l_1, r_1], [l_2, r_2], \ldots, [l_m, r_m]$. He thinks that a permutation $p$ is interesting if there exists an integer $k_i$ for every interval ($l_i \le k_i < r_i$), and if he lets $a_i = \max\limits_{j = l_i}^{k_i} p_j, b_i = \min\limits_{j = k_i + 1}^{r_i} p_j$ for every integer $i$ from $1$ to $m$, the following condition holds:

$$\max\limits_{i = 1}^m a_i < \min\limits_{i = 1}^m b_i$$

Turtle wants you to calculate the maximum number of inversions of all interesting permutations of length $n$, or tell him if there is no interesting permutation.

An inversion of a permutation $p$ is a pair of integers $(i, j)$ ($1 \le i < j \le n$) such that $p_i > p_j$.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^3$). The description of the test cases follows.

The first line of each test case contains two integers $n, m$ ($2 \le n \le 5 \cdot 10^3, 0 \le m \le 5 \cdot 10^3$) — the length of the permutation and the number of intervals.

The $i$-th of the following $m$ lines contains two integers $l_i, r_i$ ($1 \le l_i < r_i \le n$) — the $i$-th interval. Note that there may exist identical intervals (i.e., there may exist two different indices $i, j$ such that $l_i = l_j$ and $r_i = r_j$).

It is guaranteed that the sum of $n$ over all test cases does not exceed $5 \cdot 10^3$ and the sum of $m$ over all test cases does not exceed $5 \cdot 10^3$.

Output: For each test case, if there is no interesting permutation, output a single integer $-1$.

Otherwise, output a single integer — the maximum number of inversions.

Examples:
input:
8
2 0
2 1
1 2
5 1
2 4
8 3
1 4
2 5
7 8
7 2
1 4
4 7
7 3
1 2
1 7
3 7
7 4
1 3
4 7
1 3
4 7
7 3
1 2
3 4
5 6
output:
1
0
8
18
-1
-1
15
15

Note: In the third test case, the interesting permutation with the maximum number of inversions is $[5, 2, 4, 3, 1]$.

In the fourth test case, the interesting permutation with the maximum number of inversions is $[4, 3, 8, 7, 6, 2, 1, 5]$. In this case, we can let $[k_1, k_2, k_3] = [2, 2, 7]$.

In the fifth and the sixth test case, it can be proven that there is no interesting permutation.

In the seventh test case, the interesting permutation with the maximum number of inversions is $[4, 7, 6, 3, 2, 1, 5]$. In this case, we can let $[k_1, k_2, k_3, k_4] = [1, 6, 1, 6]$.

In the eighth test case, the interesting permutation with the maximum number of inversions is $[4, 7, 3, 6, 2, 5, 1]$.

Buggy Code:
#include <bits/stdc++.h>
#define forr(i,a,b) for(int i=(a);i<(b);i++)
#define forn(i,n) forr(i,0,n)
#define dforn(i,n) for(int i=n-1;i>=0;i--)
#define forall(it,v) for(auto it=v.begin();it!=v.end();it++)
#define sz(c) ((int)c.size())
#define rsz resize
#define pb push_back
#define mp make_pair
#define lb lower_bound
#define ub upper_bound
#define fst first
#define snd second

#ifdef ANARAP
//prueba
#else
//real
#endif

using namespace std;

typedef long long ll;
typedef pair<int,int> ii;

#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace __gnu_pbds;
//<key,mapped type,comparator,...>
typedef tree<int,null_type,less<int>,rb_tree_tag,
	tree_order_statistics_node_update> indexed_set;
// find_by_order(i) returns iterator to the i-th elemnt
// order_of_key(k): returns position of the lower bound of k (0-indexed)
// Ej: 12, 100, 505, 1000, 10000.
// order_of_key(10) == 0, order_of_key(100) == 1,
// order_of_key(707) == 3, order_of_key(9999999) == 5

ll contar(vector<int>& v){
	int n = sz(v);
	ll cont2 = 0;
	indexed_set os;
	forn(i,n){
		os.insert(v[i]);
		cont2 += (v[i] - os.order_of_key(v[i]));
	}
	return cont2;
}
vector<int> af;
vector<int> fijado;
ll build(vector<ii>& vq, int M, int n){
	vector<int> ret(n,-1);
	vector<int> act = fijado;
	forn(i,n){
		if(i>=M && act[i] == -1) act[i] = 0;
		else if(act[i] == -1) act[i]=1;
	}
	int cont = 0;
	dforn(i,n) if(act[i] == 0) ret[i] = cont++;
	dforn(i,n) if(act[i] == 1) ret[i] = cont++;
	ll retret = contar(ret);
	return retret;
}

int main()
{
	#ifdef ANARAP
		freopen("input.in", "r", stdin);
		//freopen("output.out", "w", stdout);
		unsigned t0, t1; t0=clock();
	#endif
	ios::sync_with_stdio(false);
	cin.tie(NULL);
	cout.tie(NULL);

	int t; cin >> t;
	forn(T,t){
		int n; cin >> n;
		int q; cin >> q;
		vector<int> v(n), vans(n, -1);
		vector<ii> vq(q);
		fijado.clear(); fijado.resize(n);
		forn(i,n) fijado[i] = -1;
		bool rip = 0;
		set<ii> s;
		forn(i,q){
			cin >> vq[i].fst >> vq[i].snd;
			vq[i].fst--; vq[i].snd--;
			s.insert(mp(vq[i].fst, vq[i].snd));
			if(fijado[vq[i].fst] == 1) rip=1;
			else fijado[vq[i].fst]=0;
			if(fijado[vq[i].snd] == 0) rip = 1;
			else fijado[vq[i].snd]=1;
		}
		while(sz(s) >= 2){
			auto it = s.begin();
			auto it2 = next(s.begin());
			if(it->snd < it2->fst){
				s.erase(it);
				continue;
			}
			else if(it->snd >= it2->snd){
				s.erase(it2);
				continue;
			}
			else{
				ii aux = *it;
				ii aux2 = *it2;
				s.erase(aux);
				s.erase(aux2);
				s.insert(mp(aux2.fst, aux.snd));
				for(int i = aux.fst; i<aux2.fst; i++){
					if(fijado[i] == 1) rip = 1;
					else fijado[i] = 0;
				}
				for(int i=aux.snd+1; i<=aux2.snd; i++){
					if(fijado[i] == 0) rip = 1;
					else fijado[i] = 1;
				}
			}
		}
		if(rip){cout << -1 << '\n'; continue;}
		ll ans = 0;
		map<int,ll> dp;
		int L=0, R=n;
		int cont = 0;
		while(L<R){
			cont++;
			if(cont>max(50, int(sqrt(n)+2))) break;
			int M1 = L+(R-L)/3;
			int M2 = R-(R-L)/3;
			//~ cout << L << " " << M1 << " " << " " << M2 << " " << R << endl;
			if(!dp.count(M1)) dp[M1] = build(vq,M1,n);
			if(!dp.count(M2)) dp[M2] = build(vq,M2,n);

			if(dp[M1] > dp[M2]) R=M2;
			else L=M1;
		}
		forall(it,dp) ans = max(it->snd, ans);
		cout << ans << '\n';
	}

	#ifdef ANARAP
	t1 = clock();double time = (double(t1-t0)/CLOCKS_PER_SEC); cout << "Execution Time: " << time << endl;
	#endif
	return 0;
}

**Instructions:**
- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.
- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.
- Output the test input directly, not code to generate it.

**Output format:**
```plaintext
{test input}
```

Think step by step.

Response:
```plaintext
1
5 3
1 3
2 4
4 5
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Shohag Loves XOR (Easy Version)
time_limit_ms: 2000
memory_limit_mb: 256
Description: This is the easy version of the problem. The differences between the two versions are highlighted in bold. You can only make hacks if both versions of the problem are solved.

Shohag has two integers $$$x$$$ and $$$m$$$. Help him count the number of integers $$$1 \le y \le m$$$ such that $$$\mathbf{x \neq y}$$$ and $$$x \oplus y$$$ is a divisor$$$^{\text{∗}}$$$ of either $$$x$$$, $$$y$$$, or both. Here $$$\oplus$$$ is the bitwise XOR operator.

Input: The first line contains a single integer $$$t$$$ ($$$1 \le t \le 10^4$$$) — the number of test cases.

The first and only line of each test case contains two space-separated integers $$$x$$$ and $$$m$$$ ($$$1 \le x \le 10^6$$$, $$$1 \le m \le 10^{18}$$$).

It is guaranteed that the sum of $$$x$$$ over all test cases does not exceed $$$10^7$$$.

Output: For each test case, print a single integer — the number of suitable $$$y$$$.

Examples:
input:
5
6 9
5 7
2 3
6 4
4 1
output:
3
2
1
1
0

Note: In the first test case, for $$$x = 6$$$, there are $$$3$$$ valid values for $$$y$$$ among the integers from $$$1$$$ to $$$m = 9$$$, and they are $$$4$$$, $$$5$$$, and $$$7$$$.

- $$$y = 4$$$ is valid because $$$x \oplus y = 6 \oplus 4 = 2$$$ and $$$2$$$ is a divisor of both $$$x = 6$$$ and $$$y = 4$$$.
- $$$y = 5$$$ is valid because $$$x \oplus y = 6 \oplus 5 = 3$$$ and $$$3$$$ is a divisor of $$$x = 6$$$.
- $$$y = 7$$$ is valid because $$$x \oplus y = 6 \oplus 7 = 1$$$ and $$$1$$$ is a divisor of both $$$x = 6$$$ and $$$y = 7$$$.

In the second test case, for $$$x = 5$$$, there are $$$2$$$ valid values for $$$y$$$ among the integers from $$$1$$$ to $$$m = 7$$$, and they are $$$4$$$ and $$$6$$$.

- $$$y = 4$$$ is valid because $$$x \oplus y = 5 \oplus 4 = 1$$$ and $$$1$$$ is a divisor of both $$$x = 5$$$ and $$$y = 4$$$.
- $$$y = 6$$$ is valid because $$$x \oplus y = 5 \oplus 6 = 3$$$ and $$$3$$$ is a divisor of $$$y = 6$$$.

Buggy Code:
#include <iostream>
using namespace std;
typedef long long LL;

int main() {
	int t;
	scanf("%d", &t);
	while (t--) {
		LL x, m, ans = 0;
		scanf("%lld%lld", &x, &m);
		for (LL i = 1; i <= (m / i) + 1; i++) {
			int y = x ^ i;
			if ((x % i == 0 || y % i == 0) && y <= m && y > 0) {
				if (x != y)
					ans++;
			}
		}
		printf("%lld\n", ans);
	}
	return 0;
}

**Instructions:**
- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.
- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.
- Output the test input directly, not code to generate it.

**Output format:**
```plaintext
{test input}
```

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided buggy code, we need to focus on edge cases and scenarios that are likely to cause the code to fail. Here are the steps to consider:

1. **Edge Cases for `x` and `m`**:
   - `x` and `m` should be chosen such that the loop condition `for (LL i = 1; i <= (m / i) + 1; i++)` might behave unexpectedly.
   - Consider values of `m` that are very large or values of `x` that are very small or large.

2. **Special Values for `x`**:
   - Values of `x` that are powers of 2 (e.g., 1, 2, 4, 8, etc.) because these values have only one bit set in their binary representation.
   - Values of `x` that are small (e.g., 1, 2, 3) to test the boundary conditions.

3. **Special Values for `m`**:
   - Values of `m` that are just slightly larger than `x` to ensure the loop runs for a few iterations.
   - Values of `m` that are very large to test the loop's performance and correctness.

4. **Values of `y`**:
   - Ensure that `y` is not equal to `x` and that `x ^ y` is a divisor of `x`, `y`, or both.

Based on these considerations, here is a challenging test input:

```plaintext
3
1 10
2 10
4 10
```

### Explanation:
- **Test Case 1: `x = 1, m = 10`**
  - `x` is 1, which is a power of 2.
  - The loop will iterate from 1 to 10, and `y` will be `1 ^ i`.
  - Valid `y` values are 2, 3, 5, 6, 7, 9, 10 (since 1 ^ 1 = 0, 1 ^ 2 = 3, 1 ^ 3 = 2, etc.).

- **Test Case 2: `x = 2, m = 10`**
  - `x` is 2, which is a power of 2.
  - The loop will iterate from 1 to 10, and `y` will be `2 ^ i`.
  - Valid `y` values are 3, 5, 7, 9 (since 2 ^ 1 = 3, 2 ^ 3 = 1, 2 ^ 5 = 7, etc.).

- **Test Case 3: `x = 4, m = 10`**
  - `x` is 4, which is a power of 2.
  - The loop will iterate from 1 to 10, and `y` will be `4 ^ i`.
  - Valid `y` values are 5, 7, 9 (since 4 ^ 1 = 5, 4 ^ 3 = 7, 4 ^ 5 = 1, etc.).

This test input should expose the bug in the provided code by testing edge cases and special values.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Shohag Loves XOR (Easy Version)
time_limit_ms: 2000
memory_limit_mb: 256
Description: This is the easy version of the problem. The differences between the two versions are highlighted in bold. You can only make hacks if both versions of the problem are solved.

Shohag has two integers $$$x$$$ and $$$m$$$. Help him count the number of integers $$$1 \le y \le m$$$ such that $$$\mathbf{x \neq y}$$$ and $$$x \oplus y$$$ is a divisor$$$^{\text{∗}}$$$ of either $$$x$$$, $$$y$$$, or both. Here $$$\oplus$$$ is the bitwise XOR operator.

Input: The first line contains a single integer $$$t$$$ ($$$1 \le t \le 10^4$$$) — the number of test cases.

The first and only line of each test case contains two space-separated integers $$$x$$$ and $$$m$$$ ($$$1 \le x \le 10^6$$$, $$$1 \le m \le 10^{18}$$$).

It is guaranteed that the sum of $$$x$$$ over all test cases does not exceed $$$10^7$$$.

Output: For each test case, print a single integer — the number of suitable $$$y$$$.

Examples:
input:
5
6 9
5 7
2 3
6 4
4 1
output:
3
2
1
1
0

Note: In the first test case, for $$$x = 6$$$, there are $$$3$$$ valid values for $$$y$$$ among the integers from $$$1$$$ to $$$m = 9$$$, and they are $$$4$$$, $$$5$$$, and $$$7$$$.

- $$$y = 4$$$ is valid because $$$x \oplus y = 6 \oplus 4 = 2$$$ and $$$2$$$ is a divisor of both $$$x = 6$$$ and $$$y = 4$$$.
- $$$y = 5$$$ is valid because $$$x \oplus y = 6 \oplus 5 = 3$$$ and $$$3$$$ is a divisor of $$$x = 6$$$.
- $$$y = 7$$$ is valid because $$$x \oplus y = 6 \oplus 7 = 1$$$ and $$$1$$$ is a divisor of both $$$x = 6$$$ and $$$y = 7$$$.

In the second test case, for $$$x = 5$$$, there are $$$2$$$ valid values for $$$y$$$ among the integers from $$$1$$$ to $$$m = 7$$$, and they are $$$4$$$ and $$$6$$$.

- $$$y = 4$$$ is valid because $$$x \oplus y = 5 \oplus 4 = 1$$$ and $$$1$$$ is a divisor of both $$$x = 5$$$ and $$$y = 4$$$.
- $$$y = 6$$$ is valid because $$$x \oplus y = 5 \oplus 6 = 3$$$ and $$$3$$$ is a divisor of $$$y = 6$$$.

Buggy Code:
#include <iostream>
#include <string>
#include <vector>
#include <cmath>
using namespace std;
int main()
{
    int t;
    cin >> t;
    while (t--)
    {
        long long x, m, res = 0;
        cin >> x >> m;
        int limit = min(2 * x, m);
        for (size_t i = floor(x * (2 / 3.0)); i <= limit; i++)
        {
            if (x != i && (x % (x ^ i) == 0 || i % (x ^ i) == 0))
                res++;
        }
        cout << res << endl;
    }
}

**Instructions:**
- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.
- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.
- Output the test input directly, not code to generate it.

**Output format:**
```plaintext
{test input}
```

Think step by step.

Response:
```plaintext
1
1 1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Shohag Loves XOR (Easy Version)
time_limit_ms: 2000
memory_limit_mb: 256
Description: This is the easy version of the problem. The differences between the two versions are highlighted in bold. You can only make hacks if both versions of the problem are solved.

Shohag has two integers $$$x$$$ and $$$m$$$. Help him count the number of integers $$$1 \le y \le m$$$ such that $$$\mathbf{x \neq y}$$$ and $$$x \oplus y$$$ is a divisor$$$^{\text{∗}}$$$ of either $$$x$$$, $$$y$$$, or both. Here $$$\oplus$$$ is the bitwise XOR operator.

Input: The first line contains a single integer $$$t$$$ ($$$1 \le t \le 10^4$$$) — the number of test cases.

The first and only line of each test case contains two space-separated integers $$$x$$$ and $$$m$$$ ($$$1 \le x \le 10^6$$$, $$$1 \le m \le 10^{18}$$$).

It is guaranteed that the sum of $$$x$$$ over all test cases does not exceed $$$10^7$$$.

Output: For each test case, print a single integer — the number of suitable $$$y$$$.

Examples:
input:
5
6 9
5 7
2 3
6 4
4 1
output:
3
2
1
1
0

Note: In the first test case, for $$$x = 6$$$, there are $$$3$$$ valid values for $$$y$$$ among the integers from $$$1$$$ to $$$m = 9$$$, and they are $$$4$$$, $$$5$$$, and $$$7$$$.

- $$$y = 4$$$ is valid because $$$x \oplus y = 6 \oplus 4 = 2$$$ and $$$2$$$ is a divisor of both $$$x = 6$$$ and $$$y = 4$$$.
- $$$y = 5$$$ is valid because $$$x \oplus y = 6 \oplus 5 = 3$$$ and $$$3$$$ is a divisor of $$$x = 6$$$.
- $$$y = 7$$$ is valid because $$$x \oplus y = 6 \oplus 7 = 1$$$ and $$$1$$$ is a divisor of both $$$x = 6$$$ and $$$y = 7$$$.

In the second test case, for $$$x = 5$$$, there are $$$2$$$ valid values for $$$y$$$ among the integers from $$$1$$$ to $$$m = 7$$$, and they are $$$4$$$ and $$$6$$$.

- $$$y = 4$$$ is valid because $$$x \oplus y = 5 \oplus 4 = 1$$$ and $$$1$$$ is a divisor of both $$$x = 5$$$ and $$$y = 4$$$.
- $$$y = 6$$$ is valid because $$$x \oplus y = 5 \oplus 6 = 3$$$ and $$$3$$$ is a divisor of $$$y = 6$$$.

Buggy Code:
#include<iostream>
using namespace std;
int check(int y,int x){
    int k= x ^ y;
    if(x%k==0 || y%k==0){
        return 1;
    }
    else{
        return 0;
    }
}
void calc(){
    int x,m;
    cin>>x>>m;
    int count=0;
    if(x<m){
    for(int y=1;y<=x+1;y++){
        if(y!=x) count+=check(y,x);

    }
    }
    else{
        for(int y=1;y<=m;y++){
        if(y!=x) count+=check(y,x);

    }
    }
    cout<<count<<endl;
}
int main(){
   int t;
   cin>>t;
   while(t--){
    calc();
   }
}

**Instructions:**
- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.
- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.
- Output the test input directly, not code to generate it.

**Output format:**
```plaintext
{test input}
```

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the given buggy code, we need to consider the following aspects:

1. **Edge Cases**: Values of `x` and `m` that are at the boundaries of the allowed ranges.
2. **Special Values**: Values of `x` that have specific properties (e.g., powers of 2, small numbers with many divisors).
3. **Large Values**: Large values of `m` to stress the algorithm.
4. **Specific Conditions**: Values that make the XOR operation result in a divisor of `x` or `y`.

Let's break it down step by step:

1. **Edge Case for `x` and `m`**:
   - `x = 1` and `m = 10^18` (to test the smallest possible `x` with a very large `m`).
   - `x = 10^6` and `m = 10^6` (to test the largest possible `x` with a large `m`).

2. **Special Values for `x`**:
   - `x = 2` and `m = 10` (to test a small `x` with a small `m`).
   - `x = 4` and `m = 10` (to test a power of 2).

3. **Large Values for `m`**:
   - `x = 5` and `m = 10^18` (to test a small `x` with a very large `m`).

4. **Specific Conditions**:
   - `x = 6` and `m = 9` (to test a case where the expected output is known and can be verified).

Combining these, we can generate a small-scale test input that is complete and valid:

```plaintext
4
1 1000000000000000000
1000000 1000000
2 10
4 10
```

This test input includes:
- A case with the smallest `x` and a very large `m`.
- A case with the largest `x` and a large `m`.
- A case with a small `x` that is a power of 2.
- A case with a small `x` and a small `m` where the expected output is known.

These cases should help expose any bugs in the given buggy code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Shohag Loves XOR (Easy Version)
time_limit_ms: 2000
memory_limit_mb: 256
Description: This is the easy version of the problem. The differences between the two versions are highlighted in bold. You can only make hacks if both versions of the problem are solved.

Shohag has two integers $$$x$$$ and $$$m$$$. Help him count the number of integers $$$1 \le y \le m$$$ such that $$$\mathbf{x \neq y}$$$ and $$$x \oplus y$$$ is a divisor$$$^{\text{∗}}$$$ of either $$$x$$$, $$$y$$$, or both. Here $$$\oplus$$$ is the bitwise XOR operator.

Input: The first line contains a single integer $$$t$$$ ($$$1 \le t \le 10^4$$$) — the number of test cases.

The first and only line of each test case contains two space-separated integers $$$x$$$ and $$$m$$$ ($$$1 \le x \le 10^6$$$, $$$1 \le m \le 10^{18}$$$).

It is guaranteed that the sum of $$$x$$$ over all test cases does not exceed $$$10^7$$$.

Output: For each test case, print a single integer — the number of suitable $$$y$$$.

Examples:
input:
5
6 9
5 7
2 3
6 4
4 1
output:
3
2
1
1
0

Note: In the first test case, for $$$x = 6$$$, there are $$$3$$$ valid values for $$$y$$$ among the integers from $$$1$$$ to $$$m = 9$$$, and they are $$$4$$$, $$$5$$$, and $$$7$$$.

- $$$y = 4$$$ is valid because $$$x \oplus y = 6 \oplus 4 = 2$$$ and $$$2$$$ is a divisor of both $$$x = 6$$$ and $$$y = 4$$$.
- $$$y = 5$$$ is valid because $$$x \oplus y = 6 \oplus 5 = 3$$$ and $$$3$$$ is a divisor of $$$x = 6$$$.
- $$$y = 7$$$ is valid because $$$x \oplus y = 6 \oplus 7 = 1$$$ and $$$1$$$ is a divisor of both $$$x = 6$$$ and $$$y = 7$$$.

In the second test case, for $$$x = 5$$$, there are $$$2$$$ valid values for $$$y$$$ among the integers from $$$1$$$ to $$$m = 7$$$, and they are $$$4$$$ and $$$6$$$.

- $$$y = 4$$$ is valid because $$$x \oplus y = 5 \oplus 4 = 1$$$ and $$$1$$$ is a divisor of both $$$x = 5$$$ and $$$y = 4$$$.
- $$$y = 6$$$ is valid because $$$x \oplus y = 5 \oplus 6 = 3$$$ and $$$3$$$ is a divisor of $$$y = 6$$$.

Buggy Code:
a=int(input())
for __ in range(a):
  b,c=map(int,input().split())
  count=0
  for i in range(c):
    if (b^i)>0 and i != b:
      if b%(b^i)==0 or c%(b^i)==0:
        count=count+1
  print(count-1)

**Instructions:**
- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.
- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.
- Output the test input directly, not code to generate it.

**Output format:**
```plaintext
{test input}
```

Think step by step.

Response:
To expose the bug in the provided buggy code, we need to focus on edge cases and scenarios where the code might fail. The main issues in the buggy code are:

1. The condition `if (b^i)>0 and i != b:` is unnecessary because `b^i` is always non-zero if `i != b`.
2. The condition `if b%(b^i)==0 or c%(b^i)==0:` is incorrect. It should check if `b^(b^i)` is a divisor of `b`, `i`, or both.
3. The loop should iterate from `1` to `c` inclusive, not from `0` to `c-1`.

Given these issues, a small-scale test input that can expose the bug is:

1. **Test Case 1**: `x = 1` and `m = 1`
   - Expected Output: `0`
   - Explanation: There are no valid `y` values because `y` must be different from `x`.

2. **Test Case 2**: `x = 3` and `m = 3`
   - Expected Output: `1`
   - Explanation: The only valid `y` is `2` because `3 ^ 2 = 1` and `1` is a divisor of `3`.

3. **Test Case 3**: `x = 4` and `m = 5`
   - Expected Output: `2`
   - Explanation: The valid `y` values are `1` and `5` because:
     - `4 ^ 1 = 5` and `5` is a divisor of `5`.
     - `4 ^ 5 = 1` and `1` is a divisor of `4`.

Here is the test input:

```plaintext
3
1 1
3 3
4 5
```