**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Min-Fund Prison (Hard)
time_limit_ms: 1500
memory_limit_mb: 256
Description: In the hard version, $$$2 \leq \sum n \leq 10^5$$$ and $$$1 \leq \sum m \leq 5 \times 10^{5}$$$

After a worker's strike organized by the Dementors asking for equal rights, the prison of Azkaban has suffered some damage. After settling the spirits, the Ministry of Magic is looking to renovate the prison to ensure that the Dementors are kept in check. The prison consists of $$$n$$$ prison cells and $$$m$$$ bi-directional corridors. The $$$i^{th}$$$ corridor is from cells $$$u_i$$$ to $$$v_i$$$. A subset of these cells $$$S$$$ is called a complex if any cell in $$$S$$$ is reachable from any other cell in $$$S$$$. Formally, a subset of cells $$$S$$$ is a complex if $$$x$$$ and $$$y$$$ are reachable from each other for all $$$x, y \in S$$$, using only cells from $$$S$$$ on the way. The funding required for a complex $$$S$$$ consisting of $$$k$$$ cells is defined as $$$k^2$$$.

As part of your Intro to Magical Interior Design course at Hogwarts, you have been tasked with designing the prison. The Ministry of Magic has asked that you divide the prison into $$$2$$$ complexes with $$$\textbf{exactly one corridor}$$$ connecting them, so that the Dementors can't organize union meetings. For this purpose, you are allowed to build bi-directional corridors. The funding required to build a corridor between any $$$2$$$ cells is $$$c$$$.

Due to budget cuts and the ongoing fight against the Death Eaters, you must find the $$$\textbf{minimum total funding}$$$ required to divide the prison as per the Ministry's requirements or $$$-1$$$ if no division is possible.

Note: The total funding is the sum of the funding required for the $$$2$$$ complexes and the corridors built. If after the division, the two complexes have $$$x$$$ and $$$y$$$ cells respectively and you have built a total of $$$a$$$ corridors, the total funding will be $$$x^2 + y^2 + c \times a$$$. Note that $$$x+y=n$$$.

Input: The first line contains one integer $$$t$$$ ($$$1 \leq t \leq 10^5$$$) — the number of test cases. Then $$$t$$$ test cases follow.

The first line of each test case consists of three integers $$$n, m$$$ and $$$c$$$ ($$$2 \leq n \leq 10^5$$$, $$$1 \leq m \leq 5 \times 10^{5}$$$, $$$1 \leq c \leq 10^9$$$)

$$$m$$$ lines follow, each consisting of 2 integers — $$$u_i, v_i$$$ indicating a corridor is present between cells $$$u_i$$$ and $$$v_i$$$ ($$$1 \leq u_i, v_i \leq n$$$, $$$u_i \neq v_i$$$)

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

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

It is guaranteed that there exists at most one corridor between any two cells.

Output: Print the $$$\textbf{minimum funding}$$$ required to divide the prison as per the Ministry's requirements or $$$-1$$$ if no division is possible.

Examples:
input:
4
4 6 5
4 3
2 3
2 4
1 2
4 1
3 1
6 6 2
1 4
2 5
3 6
1 5
3 5
6 5
6 5 7
1 4
2 5
3 6
3 5
6 5
7 5 4
1 4
3 6
3 5
6 5
2 7
output:
-1
20
25
33

Note: In the first test case of the sample input, there is no way to divide the prison according to the Ministry's requirements.

In the second test case, consider the corridor between cells $$$1$$$ and $$$5$$$ as the connection between the $$$2$$$ complexes consisting of $$$\{2, 3, 5, 6\}$$$ and $$$\{1, 4\}$$$ cells respectively. There are no new corridors built. The total funding is $$$4^2 + 2^2 = 20$$$. You can verify this is the minimum funding required.

In the third test case, build a corridor between $$$2$$$ and $$$4$$$. Consider the corridor between cells $$$1$$$ and $$$5$$$ as the connection between the $$$2$$$ complexes consisting of $$$\{3, 5, 6\}$$$ and $$$\{1, 2, 4\}$$$ cells respectively. The total funding is $$$3^2 + 3^2 + 7 \times 1 = 25$$$. You can verify this is the minimum funding required.

In the fourth test case, build a corridor between $$$2$$$ and $$$4$$$ and between $$$5$$$ and $$$7$$$. Consider the corridor between cells $$$5$$$ and $$$7$$$ as the connection between the $$$2$$$ complexes consisting of $$$\{1, 2, 4, 7\}$$$ and $$$\{3, 5, 6\}$$$ cells respectively. The total funding is $$$4^2 + 3^2 + 4 \times 2 = 33$$$. You can verify this is the minimum funding required.

Note for all test cases that there may be multiple ways to get the same funding but there is no other division which will have a more optimal minimum funding.

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

template<class T> bool cmin(T &i,T j) { return i > j ? i = j,true: false; }
template<class T> bool cmax(T &i,T j) { return i < j ? i = j,true: false; }

constexpr int nax = 100100;
bitset<nax> ks;
bitset<nax> ks2;
vector<int> adj[nax];
vector<int> tree[nax];
int t[nax];
int timer = 0;
int n,m;
ll c;
int ccs;
int dp[nax];
bool vst[nax];

void dfs(int x,int z) {
    t[x] = timer++;
    vst[x] = true;
    for (int &y: adj[x]) if (y != z) {
        if (t[y] != -1) {
            if (t[x] > t[y]) {
                dp[x]++;
                dp[y]--;
            }
        } else {
            tree[x].push_back(y);
            tree[y].push_back(x);
            dfs(y,x);
        }
    }
}

void dfstree(int x,int z) {
    assert(!vst[x]);
    vst[x] = true;
    for (int &y: tree[x]) if (y != z) {
        dfstree(y,x);
        dp[x] += dp[y];
    }
}

int sz[nax];

vector<int> lst;

void dfs_sz(int x,int z) {
    assert(!vst[x]);
    sz[x] = 1;
    vst[x] = true;
    for (int &y: tree[x]) if (y != z) {
        dfs_sz(y,x);
        sz[x] += sz[y];
    }
    if (dp[x] == 0) {
        lst.push_back(sz[x]);
    }
}

void reset() {
    ccs = timer = 0;
    for (int i=1;i<=n;i++) {
        t[i] = -1;
        dp[i] = 0;
        adj[i].clear();
        tree[i].clear();
        ks[i] = 0, ks2[i] = 0;
        sz[i] = 0;
    }
}

void solve() {
    cin >> n >> m >> c;
    reset();
    for (int i=1,x,y;i<=m;i++) {
        cin >> x >> y;
        adj[x].push_back(y);
        adj[y].push_back(x);
    }
    for (int i=1;i<=n;i++) {
        vst[i] = 0;
    }
    for (int i=1;i<=n;i++) {
        if (!vst[i]) {
            ccs++;
            dfs(i,-1);
        }
    }
    for (int i=1;i<=n;i++) {
        vst[i] = 0;
    }
    for (int i=1;i<=n;i++) {
        if (!vst[i]) {
            dfstree(i,-1);
        }
    }
    for (int i=1;i<=n;i++) {
        vst[i] = 0;
    }
    map<int,vector<vector<int>>> szs;
    for (int i=1;i<=n;i++) {
        if (!vst[i]) {
            lst.clear();
            dfs_sz(i,-1);
            assert(!lst.empty());
            szs[lst.back()].push_back(lst);
        }
    }
    ll ans = (ll)n*n;
    for (auto &[curr_size,w]: szs) {
        for (int i=1;i<=n;i++) {
            ks[i] = 0;
        }
        ks[0] = 1;
        for (auto &[other_size,occ]: szs) {
            if (other_size == curr_size) { continue; }
            int T = (int)occ.size(), p = 1;
            while(T != 0) {
                int sub = min(T,p);
                ks |= ks << (sub*other_size);
                T -= sub;
                p *= 2;
            }
        }
        vector<vector<int>> weights(3,vector<int>(1));
        for (int x=1;x+1<(int)w.size();x++) {
            weights[0].push_back(x*curr_size);
        }
        for (auto &v: w) {
            for (auto &x: v) {
                weights[1].push_back(x);
            }
        }
        sort(weights[1].begin(),weights[1].end());
        weights[1].resize(unique(weights[1].begin(),weights[1].end())-weights[1].begin());
        for (int i=1;i<=n;i++) {
            if (ks[i] == 1) {
                weights[2].push_back(i);
            }
        }
        sort(weights.begin(),weights.end(),[&](auto &x,auto &y) {
            return x.size() < y.size();
        });
        for (int &x: weights[0]) {
            for (int &y: weights[1]) {
                int tot = x+y;
                auto it = lower_bound(weights[2].begin(),weights[2].end(),n/2 - tot);
                if (it != weights[2].end()) {
                    tot += *it;
                    cmin(ans,(ll)tot*tot + (ll)(n-tot)*(n-tot));
                    tot -= *it;
                }
                if (it != weights[2].begin()) {
                    --it;
                    tot += *it;
                    cmin(ans,(ll)tot*tot + (ll)(n-tot)*(n-tot));
                    tot -= *it;
                }
            }
        }
    }
    cout << (ans == (ll)n*n ? -1LL : ans + c*(ccs-1)) << "\n";
}

int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr);
    int _t;
    cin >> _t;
    while(_t--) {
        solve();
    }
}

**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 4 10
1 2
2 3
3 4
4 5
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Min-Fund Prison (Hard)
time_limit_ms: 1500
memory_limit_mb: 256
Description: In the hard version, $$$2 \leq \sum n \leq 10^5$$$ and $$$1 \leq \sum m \leq 5 \times 10^{5}$$$

After a worker's strike organized by the Dementors asking for equal rights, the prison of Azkaban has suffered some damage. After settling the spirits, the Ministry of Magic is looking to renovate the prison to ensure that the Dementors are kept in check. The prison consists of $$$n$$$ prison cells and $$$m$$$ bi-directional corridors. The $$$i^{th}$$$ corridor is from cells $$$u_i$$$ to $$$v_i$$$. A subset of these cells $$$S$$$ is called a complex if any cell in $$$S$$$ is reachable from any other cell in $$$S$$$. Formally, a subset of cells $$$S$$$ is a complex if $$$x$$$ and $$$y$$$ are reachable from each other for all $$$x, y \in S$$$, using only cells from $$$S$$$ on the way. The funding required for a complex $$$S$$$ consisting of $$$k$$$ cells is defined as $$$k^2$$$.

As part of your Intro to Magical Interior Design course at Hogwarts, you have been tasked with designing the prison. The Ministry of Magic has asked that you divide the prison into $$$2$$$ complexes with $$$\textbf{exactly one corridor}$$$ connecting them, so that the Dementors can't organize union meetings. For this purpose, you are allowed to build bi-directional corridors. The funding required to build a corridor between any $$$2$$$ cells is $$$c$$$.

Due to budget cuts and the ongoing fight against the Death Eaters, you must find the $$$\textbf{minimum total funding}$$$ required to divide the prison as per the Ministry's requirements or $$$-1$$$ if no division is possible.

Note: The total funding is the sum of the funding required for the $$$2$$$ complexes and the corridors built. If after the division, the two complexes have $$$x$$$ and $$$y$$$ cells respectively and you have built a total of $$$a$$$ corridors, the total funding will be $$$x^2 + y^2 + c \times a$$$. Note that $$$x+y=n$$$.

Input: The first line contains one integer $$$t$$$ ($$$1 \leq t \leq 10^5$$$) — the number of test cases. Then $$$t$$$ test cases follow.

The first line of each test case consists of three integers $$$n, m$$$ and $$$c$$$ ($$$2 \leq n \leq 10^5$$$, $$$1 \leq m \leq 5 \times 10^{5}$$$, $$$1 \leq c \leq 10^9$$$)

$$$m$$$ lines follow, each consisting of 2 integers — $$$u_i, v_i$$$ indicating a corridor is present between cells $$$u_i$$$ and $$$v_i$$$ ($$$1 \leq u_i, v_i \leq n$$$, $$$u_i \neq v_i$$$)

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

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

It is guaranteed that there exists at most one corridor between any two cells.

Output: Print the $$$\textbf{minimum funding}$$$ required to divide the prison as per the Ministry's requirements or $$$-1$$$ if no division is possible.

Examples:
input:
4
4 6 5
4 3
2 3
2 4
1 2
4 1
3 1
6 6 2
1 4
2 5
3 6
1 5
3 5
6 5
6 5 7
1 4
2 5
3 6
3 5
6 5
7 5 4
1 4
3 6
3 5
6 5
2 7
output:
-1
20
25
33

Note: In the first test case of the sample input, there is no way to divide the prison according to the Ministry's requirements.

In the second test case, consider the corridor between cells $$$1$$$ and $$$5$$$ as the connection between the $$$2$$$ complexes consisting of $$$\{2, 3, 5, 6\}$$$ and $$$\{1, 4\}$$$ cells respectively. There are no new corridors built. The total funding is $$$4^2 + 2^2 = 20$$$. You can verify this is the minimum funding required.

In the third test case, build a corridor between $$$2$$$ and $$$4$$$. Consider the corridor between cells $$$1$$$ and $$$5$$$ as the connection between the $$$2$$$ complexes consisting of $$$\{3, 5, 6\}$$$ and $$$\{1, 2, 4\}$$$ cells respectively. The total funding is $$$3^2 + 3^2 + 7 \times 1 = 25$$$. You can verify this is the minimum funding required.

In the fourth test case, build a corridor between $$$2$$$ and $$$4$$$ and between $$$5$$$ and $$$7$$$. Consider the corridor between cells $$$5$$$ and $$$7$$$ as the connection between the $$$2$$$ complexes consisting of $$$\{1, 2, 4, 7\}$$$ and $$$\{3, 5, 6\}$$$ cells respectively. The total funding is $$$4^2 + 3^2 + 4 \times 2 = 33$$$. You can verify this is the minimum funding required.

Note for all test cases that there may be multiple ways to get the same funding but there is no other division which will have a more optimal minimum funding.

Buggy Code:
#include <bits/stdc++.h>
using namespace std;
int col[100005], dcc[100005];
int n, m, c, dcctot, cnt, allsz;
int T, dfn[100005], low[100005];
vector<int> g[200005];
bitset<100005> dp[2][2]; //滚动数组，0表示用过删边的操作，1表示没有
stack<int> S;
vector<int> sz[100005];
bool vis[200005];

void tarjan(int x, int fa){
	S.push(x);
	dfn[x] = low[x] = ++T;
	bool flag = false;
	for(auto it : g[x]){
		if(it == fa && !flag){ //不要与来的那条边一样，但如果两点之间有重边就行
			flag = true;
			continue;
		}
		if(!dfn[it]){
			tarjan(it, x);
			low[x] = min(low[x], low[it]);
		} else {
			low[x] = min(low[x], dfn[it]);
		}
	}
	if(dfn[x] == low[x]){
		int u = 0;
		dcctot++;
		dcc[dcctot] = 0;
		do{
			u = S.top();
			S.pop();
			col[u] = dcctot;
			dcc[dcctot]++;
		}while(u != x);
	}
}

int dfs(int x){
	int szz = dcc[x - n];
	vis[x] = 1;
	for(auto it : g[x]){
		if(!vis[it]){
			szz += dfs(it);
		}
	}
	sz[cnt].push_back(szz);
	return szz;
}
void solve(){
	while(!S.empty()) S.pop();
	cin >> n >> m >> c;
	dcctot = T = 0;
	for(int i = 1; i <= n; i++) {
		g[i].clear();
		dfn[i] = low[i] = 0;
	}
	for(int i = 1; i <= m; i++){
		int x, y;
		cin >> x >> y;
		g[x].push_back(y);
		g[y].push_back(x);
	}
	for(int i = 1; i <= n; i++) if(!dfn[i]) tarjan(i, 0);
	for(int i = n + 1; i <= 2 * n; i++) g[i].clear();
	for(int i = 1; i <= n; i++){
		for(auto it : g[i]){
			if(col[it] != col[i]){
				g[col[i] + n].push_back(n + col[it]);
			}
		}
	}
	if(dcctot == 1){ //不可能把他分成两份
		cout <<-1 << endl;
		return ;
	}
	cnt = 0;
	for(int i = 1; i <= 2 * n; i++) vis[i] = 0;
	for(int i = 1; i <= dcctot; i++){
		if(!vis[i + n]){
			sz[++cnt].clear();
			allsz = dfs(i + n);
			int len = sz[cnt].size();
			for(int j = 0; j < len; j++) sz[cnt].push_back(allsz - sz[cnt][j]);
		}
	}
	for(int i = 0; i < 2; i++){
		for(int j = 0; j < 2; j++){
			dp[i][j].set(0);
		}
	}
	dp[0][0][0] = dp[1][0][0] = 1;
	for(int i = 1; i <= cnt; i++){
		int all = sz[i].size() / 2 - 1;
		dp[0][i % 2] = (dp[0][(i % 2) ^ 1] | (dp[0][(i % 2) ^ 1] << sz[i][all])); //全选
		for(auto szz : sz[i]){ //切割子树，选择要哪个
			dp[0][i % 2] |= (dp[1][(i % 2) ^ 1] << szz);
		}
		dp[1][i % 2] = (dp[1][(i % 2) ^ 1] | (dp[1][(i % 2) ^ 1] << sz[i][all]));
	}
	long long ans = 1e18;
	for(int i = n / 2; i >= 0; i--){
		if(dp[0][cnt % 2][i] || dp[1][cnt % 2][i]){
			ans = min(ans, 1ll * (cnt - 1) * c + 1ll * i * i + 1ll * (n - i) * (n - i));
		}
	}
	cout << ans << endl;
}
int main(){
	ios::sync_with_stdio(0);
	cin.tie(0); cout.tie(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
1
5 6 1
1 2
2 3
3 4
4 5
1 5
2 4
```

### Explanation:
This test case is designed to expose potential issues in the code, especially in how it handles complex graphs and the division of the prison into two complexes. Here's a breakdown:

1. **Graph Structure**:
   - The graph is a cycle with an additional edge, forming a structure that can be challenging to divide.
   - The edges are: (1, 2), (2, 3), (3, 4), (4, 5), (1, 5), and (2, 4).

2. **Expected Behavior**:
   - The code should find a way to divide the prison into two complexes with exactly one corridor connecting them.
   - The optimal division would be to split the graph into two complexes: one with cells {1, 2, 3} and the other with cells {4, 5}, using the corridor (3, 4) as the connection.
   - The total funding required would be \(3^2 + 2^2 + 1 \times 1 = 14\).

3. **Potential Issues**:
   - The code might struggle with the additional edge (2, 4) that creates a more complex structure.
   - The code might incorrectly handle the division, leading to a higher funding cost or an incorrect division.
   - The code might not correctly identify the minimum funding required, especially if it fails to consider all possible divisions.

This test case is designed to be small but challenging, ensuring that the code's logic for graph traversal, division, and funding calculation is thoroughly tested.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Min-Fund Prison (Hard)
time_limit_ms: 1500
memory_limit_mb: 256
Description: In the hard version, $$$2 \leq \sum n \leq 10^5$$$ and $$$1 \leq \sum m \leq 5 \times 10^{5}$$$

After a worker's strike organized by the Dementors asking for equal rights, the prison of Azkaban has suffered some damage. After settling the spirits, the Ministry of Magic is looking to renovate the prison to ensure that the Dementors are kept in check. The prison consists of $$$n$$$ prison cells and $$$m$$$ bi-directional corridors. The $$$i^{th}$$$ corridor is from cells $$$u_i$$$ to $$$v_i$$$. A subset of these cells $$$S$$$ is called a complex if any cell in $$$S$$$ is reachable from any other cell in $$$S$$$. Formally, a subset of cells $$$S$$$ is a complex if $$$x$$$ and $$$y$$$ are reachable from each other for all $$$x, y \in S$$$, using only cells from $$$S$$$ on the way. The funding required for a complex $$$S$$$ consisting of $$$k$$$ cells is defined as $$$k^2$$$.

As part of your Intro to Magical Interior Design course at Hogwarts, you have been tasked with designing the prison. The Ministry of Magic has asked that you divide the prison into $$$2$$$ complexes with $$$\textbf{exactly one corridor}$$$ connecting them, so that the Dementors can't organize union meetings. For this purpose, you are allowed to build bi-directional corridors. The funding required to build a corridor between any $$$2$$$ cells is $$$c$$$.

Due to budget cuts and the ongoing fight against the Death Eaters, you must find the $$$\textbf{minimum total funding}$$$ required to divide the prison as per the Ministry's requirements or $$$-1$$$ if no division is possible.

Note: The total funding is the sum of the funding required for the $$$2$$$ complexes and the corridors built. If after the division, the two complexes have $$$x$$$ and $$$y$$$ cells respectively and you have built a total of $$$a$$$ corridors, the total funding will be $$$x^2 + y^2 + c \times a$$$. Note that $$$x+y=n$$$.

Input: The first line contains one integer $$$t$$$ ($$$1 \leq t \leq 10^5$$$) — the number of test cases. Then $$$t$$$ test cases follow.

The first line of each test case consists of three integers $$$n, m$$$ and $$$c$$$ ($$$2 \leq n \leq 10^5$$$, $$$1 \leq m \leq 5 \times 10^{5}$$$, $$$1 \leq c \leq 10^9$$$)

$$$m$$$ lines follow, each consisting of 2 integers — $$$u_i, v_i$$$ indicating a corridor is present between cells $$$u_i$$$ and $$$v_i$$$ ($$$1 \leq u_i, v_i \leq n$$$, $$$u_i \neq v_i$$$)

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

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

It is guaranteed that there exists at most one corridor between any two cells.

Output: Print the $$$\textbf{minimum funding}$$$ required to divide the prison as per the Ministry's requirements or $$$-1$$$ if no division is possible.

Examples:
input:
4
4 6 5
4 3
2 3
2 4
1 2
4 1
3 1
6 6 2
1 4
2 5
3 6
1 5
3 5
6 5
6 5 7
1 4
2 5
3 6
3 5
6 5
7 5 4
1 4
3 6
3 5
6 5
2 7
output:
-1
20
25
33

Note: In the first test case of the sample input, there is no way to divide the prison according to the Ministry's requirements.

In the second test case, consider the corridor between cells $$$1$$$ and $$$5$$$ as the connection between the $$$2$$$ complexes consisting of $$$\{2, 3, 5, 6\}$$$ and $$$\{1, 4\}$$$ cells respectively. There are no new corridors built. The total funding is $$$4^2 + 2^2 = 20$$$. You can verify this is the minimum funding required.

In the third test case, build a corridor between $$$2$$$ and $$$4$$$. Consider the corridor between cells $$$1$$$ and $$$5$$$ as the connection between the $$$2$$$ complexes consisting of $$$\{3, 5, 6\}$$$ and $$$\{1, 2, 4\}$$$ cells respectively. The total funding is $$$3^2 + 3^2 + 7 \times 1 = 25$$$. You can verify this is the minimum funding required.

In the fourth test case, build a corridor between $$$2$$$ and $$$4$$$ and between $$$5$$$ and $$$7$$$. Consider the corridor between cells $$$5$$$ and $$$7$$$ as the connection between the $$$2$$$ complexes consisting of $$$\{1, 2, 4, 7\}$$$ and $$$\{3, 5, 6\}$$$ cells respectively. The total funding is $$$4^2 + 3^2 + 4 \times 2 = 33$$$. You can verify this is the minimum funding required.

Note for all test cases that there may be multiple ways to get the same funding but there is no other division which will have a more optimal minimum funding.

Buggy Code:
/*
	www.youtube.com/YugiHackerChannel
	linktr.ee/YugiHacker
*/
#include<bits/stdc++.h>
#define el cout<<"\n"
#define f0(i,n) for(int i=0;i<n;++i)
#define f1(i,n) for(int i=1;i<=n;++i)
#define maxn 100005
using namespace std;

int n, m, c;
vector <int> a[maxn];
vector <int> vi;
long long ans;

int low[maxn], num[maxn], cnt, sz[maxn], comps;
bool has_bridge;
vector <int> bridges[maxn];
vector <int> cur_brigde;
void tarjan(int u, int p = 0)
{
    sz[u]++;
    low[u] = num[u] = ++cnt;
    int node = p != 0;
    for (int v:a[u]) if (v != p)
    {
        if (num[v]) low[u] = min(low[u], num[v]);
        else
        {
            tarjan(v, u);
            low[u] = min(low[u], low[v]);
            sz[u] += sz[v];
            if (low[v] == num[v]) {
                cur_brigde.push_back(sz[v]);
                has_bridge = 1;
            }
        }
    }
}
void knapsack(){
    sort(vi.begin(), vi.end());
    vector <int> tmp;
    int N = n;
    for (int x:vi)
    {
        tmp.push_back(x);
        if (tmp.size() >= 5 && tmp.back() == *(tmp.end() - 5))
        {
            N -= tmp.back() * 2;
            tmp.pop_back(), tmp.pop_back();
        }
    }
    vi = tmp;
    ans = 1e18;
    for (int cur_size = 1; cur_size <= n; cur_size++)
    {
        if (bridges[cur_size].size() == 0) continue;
        sort(bridges[cur_size].begin(), bridges[cur_size].end());
        bool big_size = 0;
        bitset <100001> dp;
        dp[0] = 1;
        for (int x:vi){
            if (!big_size && x == cur_size){
                big_size = 1;
                continue;
            }
            dp |= dp << x;
        }
        vector <int> diffs;
        for (int x=0; x<=(N-cur_size)/2; x++)
            if (dp[x]) diffs.push_back(N - cur_size - x - x);
        sort(diffs.begin(), diffs.end());
        vector <int> diffs2;
        for (int i=0; i<bridges[cur_size].size(); i++)
        {
            if (i == 0 || bridges[cur_size][i] != bridges[cur_size][i-1])
            {
                int w1 = bridges[cur_size][i];
                int w2 = cur_size - w1;
                diffs2.push_back(abs(w1-w2));
            }
        }
        sort(diffs2.begin(), diffs2.end());
        auto calc = [] (int diff){
            int mi = (n - diff) / 2;
            int ma = n - mi;
            return 1ll * mi * mi + 1ll * ma * ma;
        };
        for (int x:diffs2){
            auto y = lower_bound(diffs.begin(), diffs.end(), x);
            if (y != diffs.end()) ans = min(ans, calc(abs(x-*y)));
            if (y != diffs.begin()){
                --y;
                ans = min(ans, calc(abs(x-*y)));
            }
        }
    }
    bitset <100001> dp;
    for (int x:vi) dp |= dp << x;
    for (int x = N/2; x >= 0; x--) if (dp[x]){
        int diff = N - x - x;
        int mi = (n - diff) / 2;
        int ma = n - mi;
        ans = min(ans, 1ll * mi * mi + 1ll * ma * ma);
        break;
    }
    ans += 1ll * c * (comps - 1);
}
void reset(){
    vi.clear();
    cnt = comps = has_bridge = 0;
    for (int u=1; u<=n; u++)
    {
        low[u] = num[u] = sz[u] = 0;
        a[u].clear();
        bridges[u].clear();
    }
}
void solve()
{
    cin >> n >> m >> c;
    for (int i=1; i<=m; i++)
    {
        int u, v; cin >> u >> v;
        a[u].push_back(v);
        a[v].push_back(u);
    }
    for (int u=1; u<=n; u++) if (!num[u]) {
        comps++;
        cur_brigde.clear();
        tarjan(u);
        vi.push_back(sz[u]);
        for (auto x:cur_brigde) bridges[sz[u]].push_back(x);
    }
    if (vi.size() >= 2 || has_bridge) {
        knapsack();
        cout << ans;
    }
    else cout << -1;
    el;
    reset();
}
main()
{
    ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0);
    //freopen("test.inp", "r", stdin);
    int t; cin >> t;
    while (t--)
    {
        solve();
    }
}

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Cosmic Rays
time_limit_ms: 2000
memory_limit_mb: 512
Description: Given an array of integers $s_1, s_2, \ldots, s_l$, every second, cosmic rays will cause all $s_i$ such that $i=1$ or $s_i\neq s_{i-1}$ to be deleted simultaneously, and the remaining parts will be concatenated together in order to form the new array $s_1, s_2, \ldots, s_{l'}$.

Define the strength of an array as the number of seconds it takes to become empty.

You are given an array of integers compressed in the form of $n$ pairs that describe the array left to right. Each pair $(a_i,b_i)$ represents $a_i$ copies of $b_i$, i.e. $\underbrace{b_i,b_i,\cdots,b_i}_{a_i\textrm{ times}}$.

For each $i=1,2,\dots,n$, please find the strength of the sequence described by the first $i$ pairs.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1\le t\le10^4$). The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($1\le n\le3\cdot10^5$) — the length of sequence $a$.

The next $n$ lines contain two integers each $a_i$, $b_i$ ($1\le a_i\le10^9,0\le b_i\le n$) — the pairs which describe the sequence.

It is guaranteed that the sum of all $n$ does not exceed $3\cdot10^5$.

It is guaranteed that for all $1\le i<n$, $b_i\neq b_{i+1}$ holds.

Output: For each test case, print one line containing $n$ integers — the answer for each prefix of pairs.

Examples:
input:
4
4
2 0
1 1
3 0
5 1
6
4 6
1 3
4 6
4 0
7 6
6 3
7
9 0
7 1
5 0
7 1
9 0
1 1
2 0
10
10 7
4 9
2 2
7 9
2 8
8 5
11 7
15 5
12 7
4 0
output:
2 2 4 5
4 4 7 7 10 10
9 9 9 9 9 9 10
10 10 10 10 10 10 12 15 15 15

Note: In the first test case, for the prefix of length $4$, the changes will be $[0,0,1,0,0,0,1,1,1,1,1]\rightarrow[0,0,0,1,1,1,1]\rightarrow[0,0,1,1,1]\rightarrow[0,1,1]\rightarrow[1]\rightarrow[]$, so the array becomes empty after $5$ seconds.

In the second test case, for the prefix of length $4$, the changes will be $[6,6,6,6,3,6,6,6,6,0,0,0,0]\rightarrow[6,6,6,6,6,6,0,0,0]\rightarrow[6,6,6,6,6,0,0]\rightarrow[6,6,6,6,0]\rightarrow[6,6,6]\rightarrow[6,6]\rightarrow[6]\rightarrow[]$, so the array becomes empty after $7$ seconds.

Buggy Code:
#include<cstdio>
#include<iostream>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<queue>
#include<array>
#include<stack>
#include<set>
#include<map>
#include<vector>
#include<ctime>
#include<bitset>
#include<tuple>
#define ull unsigned long long
#define ll long long
#define int long long
#define max_ll 9223372036854775807
#define max_int 2147483647
using namespace std;
template<class T>inline void read(T &x)
{
    x=0;char c=getchar();bool f=0;
    while(!isdigit(c))f^=c=='-',c=getchar();
    while(isdigit(c))x=(x<<3)+(x<<1)+(c^48),c=getchar();
    if(f)x=-x;
}
template<class T>inline void print(T x)
{
    if(x<0)putchar('-'),x=-x;
    if(x>9)print(x/10);
    putchar('0'+x%10);
}
stack<pair<int,int> > s;
int t;
int n;
int ans=0;
int x;int y;
signed main(){
	read(t);
	while(t--){
		read(n);
		for(int i=1;i<=n;++i){
			read(x);read(y);
			int cnt=0;
			while(!s.empty()&&(s.top().second!=y&&s.top().first<=x)){
				cnt=max(cnt,s.top().first);
				s.pop();
			}
			if(s.empty()){
				s.push({x,y});
				ans=x;goto A;
			}
			if(s.top().second==y){
				auto k=s.top();s.pop();
				k.first+=max(x-cnt,0ll);
				ans=max(ans,k.first);
				s.push(k);goto A;
			}
			s.push({x,y});
			A:;
			printf("%lld ",ans);
		}
		puts("");
		while(!s.empty())
			s.pop();
		ans=0;
	}
	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
3
1 0
1 1
1 0
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Cosmic Rays
time_limit_ms: 2000
memory_limit_mb: 512
Description: Given an array of integers $s_1, s_2, \ldots, s_l$, every second, cosmic rays will cause all $s_i$ such that $i=1$ or $s_i\neq s_{i-1}$ to be deleted simultaneously, and the remaining parts will be concatenated together in order to form the new array $s_1, s_2, \ldots, s_{l'}$.

Define the strength of an array as the number of seconds it takes to become empty.

You are given an array of integers compressed in the form of $n$ pairs that describe the array left to right. Each pair $(a_i,b_i)$ represents $a_i$ copies of $b_i$, i.e. $\underbrace{b_i,b_i,\cdots,b_i}_{a_i\textrm{ times}}$.

For each $i=1,2,\dots,n$, please find the strength of the sequence described by the first $i$ pairs.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1\le t\le10^4$). The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($1\le n\le3\cdot10^5$) — the length of sequence $a$.

The next $n$ lines contain two integers each $a_i$, $b_i$ ($1\le a_i\le10^9,0\le b_i\le n$) — the pairs which describe the sequence.

It is guaranteed that the sum of all $n$ does not exceed $3\cdot10^5$.

It is guaranteed that for all $1\le i<n$, $b_i\neq b_{i+1}$ holds.

Output: For each test case, print one line containing $n$ integers — the answer for each prefix of pairs.

Examples:
input:
4
4
2 0
1 1
3 0
5 1
6
4 6
1 3
4 6
4 0
7 6
6 3
7
9 0
7 1
5 0
7 1
9 0
1 1
2 0
10
10 7
4 9
2 2
7 9
2 8
8 5
11 7
15 5
12 7
4 0
output:
2 2 4 5
4 4 7 7 10 10
9 9 9 9 9 9 10
10 10 10 10 10 10 12 15 15 15

Note: In the first test case, for the prefix of length $4$, the changes will be $[0,0,1,0,0,0,1,1,1,1,1]\rightarrow[0,0,0,1,1,1,1]\rightarrow[0,0,1,1,1]\rightarrow[0,1,1]\rightarrow[1]\rightarrow[]$, so the array becomes empty after $5$ seconds.

In the second test case, for the prefix of length $4$, the changes will be $[6,6,6,6,3,6,6,6,6,0,0,0,0]\rightarrow[6,6,6,6,6,6,0,0,0]\rightarrow[6,6,6,6,6,0,0]\rightarrow[6,6,6,6,0]\rightarrow[6,6,6]\rightarrow[6,6]\rightarrow[6]\rightarrow[]$, so the array becomes empty after $7$ seconds.

Buggy Code:
#ifndef LOCAL
#define FAST_IO
#endif

// ============
#include <bits/stdc++.h>
#define OVERRIDE(a, b, c, d, ...) d
#define REP2(i, n) for (i32 i = 0; i < (i32)(n); ++i)
#define REP3(i, m, n) for (i32 i = (i32)(m); i < (i32)(n); ++i)
#define REP(...) OVERRIDE(__VA_ARGS__, REP3, REP2)(__VA_ARGS__)
#define PER2(i, n) for (i32 i = (i32)(n)-1; i >= 0; --i)
#define PER3(i, m, n) for (i32 i = (i32)(n)-1; i >= (i32)(m); --i)
#define PER(...) OVERRIDE(__VA_ARGS__, PER3, PER2)(__VA_ARGS__)
#define ALL(x) begin(x), end(x)
#define LEN(x) (i32)(x.size())
using namespace std;
using u32 = unsigned int;
using u64 = unsigned long long;
using i32 = signed int;
using i64 = signed long long;
using f64 = double;
using f80 = long double;
using pi = pair<i32, i32>;
using pl = pair<i64, i64>;
template <typename T>
using V = vector<T>;
template <typename T>
using VV = V<V<T>>;
template <typename T>
using VVV = V<V<V<T>>>;
template <typename T>
using VVVV = V<V<V<V<T>>>>;
template <typename T>
using PQR = priority_queue<T, V<T>, greater<T>>;
template <typename T>
bool chmin(T &x, const T &y) {
    if (x > y) {
        x = y;
        return true;
    }
    return false;
}
template <typename T>
bool chmax(T &x, const T &y) {
    if (x < y) {
        x = y;
        return true;
    }
    return false;
}
template <typename T>
i32 lob(const V<T> &arr, const T &v) {
    return (i32)(lower_bound(ALL(arr), v) - arr.begin());
}
template <typename T>
i32 upb(const V<T> &arr, const T &v) {
    return (i32)(upper_bound(ALL(arr), v) - arr.begin());
}
template <typename T>
V<i32> argsort(const V<T> &arr) {
    V<i32> ret(arr.size());
    iota(ALL(ret), 0);
    sort(ALL(ret), [&](i32 i, i32 j) -> bool {
        if (arr[i] == arr[j]) {
            return i < j;
        } else {
            return arr[i] < arr[j];
        }
    });
    return ret;
}
#ifdef INT128
using u128 = __uint128_t;
using i128 = __int128_t;
#endif
[[maybe_unused]] constexpr i32 INF = 1000000100;
[[maybe_unused]] constexpr i64 INF64 = 3000000000000000100;
struct SetUpIO {
    SetUpIO() {
#ifdef FAST_IO
        ios::sync_with_stdio(false);
        cin.tie(nullptr);
#endif
        cout << fixed << setprecision(15);
    }
} set_up_io;
void scan(char &x) { cin >> x; }
void scan(u32 &x) { cin >> x; }
void scan(u64 &x) { cin >> x; }
void scan(i32 &x) { cin >> x; }
void scan(i64 &x) { cin >> x; }
void scan(string &x) { cin >> x; }
template <typename T>
void scan(V<T> &x) {
    for (T &ele : x) {
        scan(ele);
    }
}
void read() {}
template <typename Head, typename... Tail>
void read(Head &head, Tail &...tail) {
    scan(head);
    read(tail...);
}
#define CHAR(...)     \
    char __VA_ARGS__; \
    read(__VA_ARGS__);
#define U32(...)     \
    u32 __VA_ARGS__; \
    read(__VA_ARGS__);
#define U64(...)     \
    u64 __VA_ARGS__; \
    read(__VA_ARGS__);
#define I32(...)     \
    i32 __VA_ARGS__; \
    read(__VA_ARGS__);
#define I64(...)     \
    i64 __VA_ARGS__; \
    read(__VA_ARGS__);
#define STR(...)        \
    string __VA_ARGS__; \
    read(__VA_ARGS__);
#define VEC(type, name, size) \
    V<type> name(size);       \
    read(name);
#define VVEC(type, name, size1, size2)    \
    VV<type> name(size1, V<type>(size2)); \
    read(name);
// ============

#ifdef DEBUGF
#else
#define DBG(...) (void)0
#endif

struct Node {
    i64 a;
    i32 b;
    Node *p1, *p2;
    i64 length;
    Node() : a(0), b(0), p1(nullptr), p2(nullptr), length(0) {}
    void upd() {
        length = a;
        if (p1) {
            length += p1->length;
        }
    }
};

void solve() {
    I32(n);
    V<i64> a(n);
    V<i32> b(n);
    REP(i, n) {
        read(a[i], b[i]);
    }
    Node *stc = new Node;
    V<i64> wr(n);
    REP(i, n) {
        i64 curl = a[i];
        Node *cur = stc;
        i64 t = 0;
        while (cur) {
            DBG(curl, t, cur->a, cur->b);
            if (cur->b == b[i]) {
                curl += cur->a;
                cur = cur->p2;
            } else if (cur->a <= curl) {
                t += cur->a;
                curl -= cur->a;
                cur = cur->p1;
            } else {
                break;
            }
        }
        DBG(t, curl);
        i64 ans = t;
        if (cur) {
            ans += cur->length;
        } else {
            ans += curl;
        }
        wr[i] = ans;
        if (cur) {
            Node *p1 = new Node;
            p1->a = cur->a - curl;
            p1->b = cur->b;
            p1->p1 = cur->p1;
            p1->p2 = cur->p2;
            i64 red = curl;
            while (p1->p2 && p1->p2->a <= red) {
                red -= p1->p2->a;
                p1->p2 = p1->p2->p1;
            }
            if (p1->p2) {
                Node *tmp = new Node;
                tmp->a = p1->p2->a - red;
                tmp->b = p1->p2->b;
                tmp->p1 = p1->p2->p1;
                tmp->p2 = p1->p2->p2;
                p1->p2 = tmp;
                p1->p2->upd();
            }
            p1->upd();
            Node *p2 = new Node;
            p2->a = cur->a;
            p2->b = cur->b;
            p2->p1 = cur->p1;
            p2->p2 = cur->p2;
            Node *nw = new Node;
            nw->a = t + curl;
            nw->b = b[i];
            nw->p1 = p1;
            nw->p2 = p2;
            nw->upd();
            cur = nw;
        } else {
            cur = new Node;
            cur->a = ans;
            cur->b = b[i];
            cur->upd();
        }
        stc = cur;
        DBG(stc->a, stc->b);
        if (stc->p2) {
            DBG(stc->p2->a, stc->p2->b);
        }
    }
    REP(i, n) {
        cout << wr[i] << " \n"[i + 1 == n];
    }
}

int main() {
    i32 t = 1;
    cin >> t;
    while (t--) {
        solve();
    }
}

**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 and scenarios that can highlight issues in the logic. The code is designed to handle a sequence of pairs and compute the strength of the sequence. The strength is the number of seconds it takes for the sequence to become empty under the given rules.

### Key Points to Consider:
1. **Edge Cases for Single Pairs:**
   - Test with a single pair.
   - Test with a pair where the count is 1.

2. **Edge Cases for Multiple Pairs:**
   - Test with pairs where all values are the same.
   - Test with pairs where all values are different.
   - Test with pairs that have alternating values.
   - Test with pairs that have a large count for one value and a small count for another.

3. **Edge Cases for Complex Patterns:**
   - Test with patterns that require multiple steps to become empty.
   - Test with patterns that have a mix of large and small counts.

### Chosen Test Input:
Let's create a test input that includes a mix of these scenarios to ensure the code is thoroughly tested.

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

### Explanation:
1. **Single Pair:**
   - `1 0`: This should be empty in 1 second.
   - Expected output: `1`

2. **Two Pairs with Different Values:**
   - `1 0`
   - `1 1`: This should be empty in 2 seconds.
   - Expected output: `1 2`

3. **Four Pairs with a Complex Pattern:**
   - `2 0`: Two zeros.
   - `1 1`: One one.
   - `3 0`: Three zeros.
   - `5 1`: Five ones.
   - This sequence will take 5 seconds to become empty.
   - Expected output: `2 2 4 5`

### Final Test Input:
```plaintext
3
1
1 0
2
1 0
1 1
4
2 0
1 1
3 0
5 1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Cosmic Rays
time_limit_ms: 2000
memory_limit_mb: 512
Description: Given an array of integers $s_1, s_2, \ldots, s_l$, every second, cosmic rays will cause all $s_i$ such that $i=1$ or $s_i\neq s_{i-1}$ to be deleted simultaneously, and the remaining parts will be concatenated together in order to form the new array $s_1, s_2, \ldots, s_{l'}$.

Define the strength of an array as the number of seconds it takes to become empty.

You are given an array of integers compressed in the form of $n$ pairs that describe the array left to right. Each pair $(a_i,b_i)$ represents $a_i$ copies of $b_i$, i.e. $\underbrace{b_i,b_i,\cdots,b_i}_{a_i\textrm{ times}}$.

For each $i=1,2,\dots,n$, please find the strength of the sequence described by the first $i$ pairs.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1\le t\le10^4$). The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($1\le n\le3\cdot10^5$) — the length of sequence $a$.

The next $n$ lines contain two integers each $a_i$, $b_i$ ($1\le a_i\le10^9,0\le b_i\le n$) — the pairs which describe the sequence.

It is guaranteed that the sum of all $n$ does not exceed $3\cdot10^5$.

It is guaranteed that for all $1\le i<n$, $b_i\neq b_{i+1}$ holds.

Output: For each test case, print one line containing $n$ integers — the answer for each prefix of pairs.

Examples:
input:
4
4
2 0
1 1
3 0
5 1
6
4 6
1 3
4 6
4 0
7 6
6 3
7
9 0
7 1
5 0
7 1
9 0
1 1
2 0
10
10 7
4 9
2 2
7 9
2 8
8 5
11 7
15 5
12 7
4 0
output:
2 2 4 5
4 4 7 7 10 10
9 9 9 9 9 9 10
10 10 10 10 10 10 12 15 15 15

Note: In the first test case, for the prefix of length $4$, the changes will be $[0,0,1,0,0,0,1,1,1,1,1]\rightarrow[0,0,0,1,1,1,1]\rightarrow[0,0,1,1,1]\rightarrow[0,1,1]\rightarrow[1]\rightarrow[]$, so the array becomes empty after $5$ seconds.

In the second test case, for the prefix of length $4$, the changes will be $[6,6,6,6,3,6,6,6,6,0,0,0,0]\rightarrow[6,6,6,6,6,6,0,0,0]\rightarrow[6,6,6,6,6,0,0]\rightarrow[6,6,6,6,0]\rightarrow[6,6,6]\rightarrow[6,6]\rightarrow[6]\rightarrow[]$, so the array becomes empty after $7$ seconds.

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

int simulate_deletion(vector<pair<int, int>>& sequence) {
    int strength = 0;

    while (!sequence.empty()) {
        strength++;
        vector<pair<int, int>> new_sequence;

        for (int i = 0; i < sequence.size(); i++) {
            if (i == 0 || sequence[i].second != sequence[i - 1].second) {
                if (sequence[i].first > 1) {
                    new_sequence.push_back({sequence[i].first - 1, sequence[i].second});
                }
            } else {
                new_sequence.push_back(sequence[i]);
            }
        }

        sequence = new_sequence;
    }

    return strength;
}

int main() {
    ios::sync_with_stdio(false);
    cin.tie(0);

    int t;
    cin >> t;

    while (t--) {
        int n;
        cin >> n;

        vector<pair<int, int>> sequence;
        vector<int> result;

        for (int i = 0; i < n; ++i) {
            int a, b;
            cin >> a >> b;
            sequence.push_back({a, b});

            vector<pair<int, int>> current_sequence(sequence.begin(), sequence.end());
            int current_strength = simulate_deletion(current_sequence);
            result.push_back(current_strength);
        }

        for (int r : result) {
            cout << r << " ";
        }
        cout << endl;
    }

    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
1 0
1 1
1 0
4
1 0
1 1
1 2
1 1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Cosmic Rays
time_limit_ms: 2000
memory_limit_mb: 512
Description: Given an array of integers $s_1, s_2, \ldots, s_l$, every second, cosmic rays will cause all $s_i$ such that $i=1$ or $s_i\neq s_{i-1}$ to be deleted simultaneously, and the remaining parts will be concatenated together in order to form the new array $s_1, s_2, \ldots, s_{l'}$.

Define the strength of an array as the number of seconds it takes to become empty.

You are given an array of integers compressed in the form of $n$ pairs that describe the array left to right. Each pair $(a_i,b_i)$ represents $a_i$ copies of $b_i$, i.e. $\underbrace{b_i,b_i,\cdots,b_i}_{a_i\textrm{ times}}$.

For each $i=1,2,\dots,n$, please find the strength of the sequence described by the first $i$ pairs.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1\le t\le10^4$). The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($1\le n\le3\cdot10^5$) — the length of sequence $a$.

The next $n$ lines contain two integers each $a_i$, $b_i$ ($1\le a_i\le10^9,0\le b_i\le n$) — the pairs which describe the sequence.

It is guaranteed that the sum of all $n$ does not exceed $3\cdot10^5$.

It is guaranteed that for all $1\le i<n$, $b_i\neq b_{i+1}$ holds.

Output: For each test case, print one line containing $n$ integers — the answer for each prefix of pairs.

Examples:
input:
4
4
2 0
1 1
3 0
5 1
6
4 6
1 3
4 6
4 0
7 6
6 3
7
9 0
7 1
5 0
7 1
9 0
1 1
2 0
10
10 7
4 9
2 2
7 9
2 8
8 5
11 7
15 5
12 7
4 0
output:
2 2 4 5
4 4 7 7 10 10
9 9 9 9 9 9 10
10 10 10 10 10 10 12 15 15 15

Note: In the first test case, for the prefix of length $4$, the changes will be $[0,0,1,0,0,0,1,1,1,1,1]\rightarrow[0,0,0,1,1,1,1]\rightarrow[0,0,1,1,1]\rightarrow[0,1,1]\rightarrow[1]\rightarrow[]$, so the array becomes empty after $5$ seconds.

In the second test case, for the prefix of length $4$, the changes will be $[6,6,6,6,3,6,6,6,6,0,0,0,0]\rightarrow[6,6,6,6,6,6,0,0,0]\rightarrow[6,6,6,6,6,0,0]\rightarrow[6,6,6,6,0]\rightarrow[6,6,6]\rightarrow[6,6]\rightarrow[6]\rightarrow[]$, so the array becomes empty after $7$ seconds.

Buggy Code:
t = int(input())
for i in range (t):
    n = int(input())
    a=[0]*n
    b=[0]*n
    for i in range(n):
        a[i],b[i] = (int(k) for k in input().split())

    print(a[0],end = " ")
    info = [[a[0],b[0]],]
    strength = a[0]
    maxm_idx = 0
    for i in range(1,n):
        # if (i==6):
        #     print("")
        if (len(info)==1):
            if (strength < a[i]):
                strength = a[i]
                info = [[a[i],b[i]],]
                maxm_idx = i
                print(strength,end=" ")
                continue
            else:
                info.append([a[i],b[i]])
        else:
            if (a[i]>info[len(info)-1][0]):
                # info.append([a[i]-(info[len(info)-1][0]),b[i]])
                info.append([a[i],b[i]])
                # z = info[len(info)-2][0]
                z=0
                # info.pop(-2)
                offset = 0
                num_elem = 0
                for j in range(len(info)-2,maxm_idx-1,-1):
                    if (a[i]+offset<0):
                        break
                    if (info[j][1]!=b[i]):
                        # offset -= info[j][0]
                        if (info[j][0]>offset):
                            offset = -1*info[j][0]
                        continue
                    else:
                        info[j][0] += a[i]+offset
                        info = info[:j+1]
                        break
                for l in range(len(info)):
                    if info[l][0] > strength:
                        strength = info[l][0]
                        maxm_idx = l
            else:
                info.append([a[i],b[i]])
        print(strength,end=" ")

**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
1 0
1 1
1 0
4
1 0
1 1
1 0
1 1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Cosmic Rays
time_limit_ms: 2000
memory_limit_mb: 512
Description: Given an array of integers $s_1, s_2, \ldots, s_l$, every second, cosmic rays will cause all $s_i$ such that $i=1$ or $s_i\neq s_{i-1}$ to be deleted simultaneously, and the remaining parts will be concatenated together in order to form the new array $s_1, s_2, \ldots, s_{l'}$.

Define the strength of an array as the number of seconds it takes to become empty.

You are given an array of integers compressed in the form of $n$ pairs that describe the array left to right. Each pair $(a_i,b_i)$ represents $a_i$ copies of $b_i$, i.e. $\underbrace{b_i,b_i,\cdots,b_i}_{a_i\textrm{ times}}$.

For each $i=1,2,\dots,n$, please find the strength of the sequence described by the first $i$ pairs.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1\le t\le10^4$). The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($1\le n\le3\cdot10^5$) — the length of sequence $a$.

The next $n$ lines contain two integers each $a_i$, $b_i$ ($1\le a_i\le10^9,0\le b_i\le n$) — the pairs which describe the sequence.

It is guaranteed that the sum of all $n$ does not exceed $3\cdot10^5$.

It is guaranteed that for all $1\le i<n$, $b_i\neq b_{i+1}$ holds.

Output: For each test case, print one line containing $n$ integers — the answer for each prefix of pairs.

Examples:
input:
4
4
2 0
1 1
3 0
5 1
6
4 6
1 3
4 6
4 0
7 6
6 3
7
9 0
7 1
5 0
7 1
9 0
1 1
2 0
10
10 7
4 9
2 2
7 9
2 8
8 5
11 7
15 5
12 7
4 0
output:
2 2 4 5
4 4 7 7 10 10
9 9 9 9 9 9 10
10 10 10 10 10 10 12 15 15 15

Note: In the first test case, for the prefix of length $4$, the changes will be $[0,0,1,0,0,0,1,1,1,1,1]\rightarrow[0,0,0,1,1,1,1]\rightarrow[0,0,1,1,1]\rightarrow[0,1,1]\rightarrow[1]\rightarrow[]$, so the array becomes empty after $5$ seconds.

In the second test case, for the prefix of length $4$, the changes will be $[6,6,6,6,3,6,6,6,6,0,0,0,0]\rightarrow[6,6,6,6,6,6,0,0,0]\rightarrow[6,6,6,6,6,0,0]\rightarrow[6,6,6,6,0]\rightarrow[6,6,6]\rightarrow[6,6]\rightarrow[6]\rightarrow[]$, so the array becomes empty after $7$ seconds.

Buggy Code:
from collections import defaultdict,Counter,deque
import math
from functools import lru_cache
import random
from sys import stdin, stdout
import itertools
import heapq
from fractions import Fraction
import string

def read():
    return stdin.readline().rstrip()

def write(*args, **kwargs):
    sep = kwargs.get('sep', ' ')
    end = kwargs.get('end', '\n')
    stdout.write(sep.join(str(a) for a in args) + end)

#
mod = 998244353
# mod = 10**9+7
# N = 200007
# fct = [1]*N
# for i in range(1, N):
#     fct[i] = fct[i-1] * i % mod
# ifct = fct[:]
# ifct[-1] = pow(fct[-1], mod-2, mod)
# for i in reversed(range(N-1)):
#     ifct[i] = ifct[i+1] * (i+1) % mod

# def C(n, r):
#     if n < r or r < 0: return 0
#     return fct[n] * ifct[r] % mod * ifct[n-r] % mod

total = int(input())
# total = 1

for _ in range(total):
    n = int(input())
    # n, m, k =  [*map(int, read().split(' '))]
    # a =   [*map(int, read().split(' '))]
    # b =   [*map(int, read().split(' '))]

    # if a==b or a == b[::-1]:
    #     print('Bob')
    # else:
    #     print('Alice')\

    sm = 0
    ret = []
    st = []
    mx = 0
    for i in range(n):
        a,b =  [*map(int, read().split(' '))]
        z = 0
        while st and a >= st[-1][0] and b!= st[-1][1]:
            z = max(z, st[-1][0])
            st.pop()
        if st and    b== st[-1][1]:
            st[-1][0] += a-z
        else:
            st.append([a,b])
        mx = max(mx, st[-1][0])
        ret.append(mx)

    print(*ret)

    # print(*ret)

**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 test input that exposes the bug in the provided buggy code, we need to focus on edge cases and scenarios that can maximize the failure probability. The buggy code seems to have issues with handling sequences where the same values are repeated in a way that could cause unexpected behavior in the stack operations.

### Step-by-Step Reasoning:

1. **Understanding the Buggy Code**:
   - The code uses a stack to keep track of the counts and values of the sequence.
   - It pops elements from the stack when the current element's count is greater than or equal to the top of the stack and the values are different.
   - It then updates the stack and calculates the maximum count seen so far.

2. **Identifying Potential Issues**:
   - The code might have issues when the same value is repeated multiple times, especially if the counts are not handled correctly.
   - The stack operations might not correctly handle the removal of elements, leading to incorrect maximum counts.

3. **Constructing a Test Case**:
   - We need a sequence that causes the stack to be manipulated in a way that exposes the bug.
   - A sequence with repeated values and varying counts can be a good candidate.

### Test Input:
```plaintext
1
3
1 0
2 0
1 0
```

### Explanation:
- **Test Case**:
  - `n = 3`
  - The sequence is described by the pairs: `(1, 0)`, `(2, 0)`, `(1, 0)`
  - This means the sequence is `[0, 0, 0, 0]`.

- **Expected Output**:
  - For the prefix of length 1: `[0]` → `[0]` → `[]` (strength = 2)
  - For the prefix of length 2: `[0, 0, 0]` → `[0, 0]` → `[0]` → `[]` (strength = 3)
  - For the prefix of length 3: `[0, 0, 0, 0]` → `[0, 0, 0]` → `[0, 0]` → `[0]` → `[]` (strength = 4)

- **Output**:
  - `2 3 4`

### Why This Test Case is Challenging:
- The sequence has repeated values (`0`), and the counts are such that the stack operations in the buggy code might not handle them correctly.
- The expected output requires the code to correctly manage the stack and update the maximum count, which can expose any issues in the stack manipulation logic.

This test case should help in identifying and fixing the bug 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: Cosmic Rays
time_limit_ms: 2000
memory_limit_mb: 512
Description: Given an array of integers $s_1, s_2, \ldots, s_l$, every second, cosmic rays will cause all $s_i$ such that $i=1$ or $s_i\neq s_{i-1}$ to be deleted simultaneously, and the remaining parts will be concatenated together in order to form the new array $s_1, s_2, \ldots, s_{l'}$.

Define the strength of an array as the number of seconds it takes to become empty.

You are given an array of integers compressed in the form of $n$ pairs that describe the array left to right. Each pair $(a_i,b_i)$ represents $a_i$ copies of $b_i$, i.e. $\underbrace{b_i,b_i,\cdots,b_i}_{a_i\textrm{ times}}$.

For each $i=1,2,\dots,n$, please find the strength of the sequence described by the first $i$ pairs.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1\le t\le10^4$). The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($1\le n\le3\cdot10^5$) — the length of sequence $a$.

The next $n$ lines contain two integers each $a_i$, $b_i$ ($1\le a_i\le10^9,0\le b_i\le n$) — the pairs which describe the sequence.

It is guaranteed that the sum of all $n$ does not exceed $3\cdot10^5$.

It is guaranteed that for all $1\le i<n$, $b_i\neq b_{i+1}$ holds.

Output: For each test case, print one line containing $n$ integers — the answer for each prefix of pairs.

Examples:
input:
4
4
2 0
1 1
3 0
5 1
6
4 6
1 3
4 6
4 0
7 6
6 3
7
9 0
7 1
5 0
7 1
9 0
1 1
2 0
10
10 7
4 9
2 2
7 9
2 8
8 5
11 7
15 5
12 7
4 0
output:
2 2 4 5
4 4 7 7 10 10
9 9 9 9 9 9 10
10 10 10 10 10 10 12 15 15 15

Note: In the first test case, for the prefix of length $4$, the changes will be $[0,0,1,0,0,0,1,1,1,1,1]\rightarrow[0,0,0,1,1,1,1]\rightarrow[0,0,1,1,1]\rightarrow[0,1,1]\rightarrow[1]\rightarrow[]$, so the array becomes empty after $5$ seconds.

In the second test case, for the prefix of length $4$, the changes will be $[6,6,6,6,3,6,6,6,6,0,0,0,0]\rightarrow[6,6,6,6,6,6,0,0,0]\rightarrow[6,6,6,6,6,0,0]\rightarrow[6,6,6,6,0]\rightarrow[6,6,6]\rightarrow[6,6]\rightarrow[6]\rightarrow[]$, so the array becomes empty after $7$ seconds.

Buggy Code:
import os, sys
from io import BytesIO, IOBase
from array import array
from itertools import accumulate
import bisect
import math
from collections import deque

# from functools import cache
# cache cf需要自己提交 pypy3.9!
from copy import deepcopy

class FastIO(IOBase):
    newlines = 0

    def __init__(self, file):
        self._fd = file.fileno()
        self.buffer = BytesIO()
        self.writable = "x" in file.mode or "r" not in file.mode
        self.write = self.buffer.write if self.writable else None

    def read(self):
        while True:
            b = os.read(self._fd, max(os.fstat(self._fd).st_size, 8192))
            if not b:
                break
            ptr = self.buffer.tell()
            self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
        self.newlines = 0
        return self.buffer.read()

    def readline(self):
        while self.newlines == 0:
            b = os.read(self._fd, max(os.fstat(self._fd).st_size, 8192))
            self.newlines = b.count(b"\n") + (not b)
            ptr = self.buffer.tell()
            self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
        self.newlines -= 1
        return self.buffer.readline()

    def flush(self):
        if self.writable:
            os.write(self._fd, self.buffer.getvalue())
            self.buffer.truncate(0), self.buffer.seek(0)

class IOWrapper(IOBase):
    def __init__(self, file):
        self.buffer = FastIO(file)
        self.flush = self.buffer.flush
        self.writable = self.buffer.writable
        self.write = lambda s: self.buffer.write(s.encode("ascii"))
        self.read = lambda: self.buffer.read().decode("ascii")
        self.readline = lambda: self.buffer.readline().decode("ascii")

sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)

input = lambda: sys.stdin.readline().strip()
ints = lambda: list(map(int, input().split()))
Int = lambda: int(input())

def queryInteractive(a, b, c):
    print('? {} {} {}'.format(a, b, c))
    sys.stdout.flush()
    return int(input())

def answerInteractive(x1, x2):
    print('! {} {}'.format(x1, x2))
    sys.stdout.flush()

inf = float('inf')
#if "PyPy" in sys.version:
#    import pypyjit; pypyjit.set_param('max_unroll_recursion=-1')

def solve():
    pass

T = Int()
while T:
    T -= 1
    n = Int()
    stack = []
    a0,b0 = ints()
    ans = [a0]
    stack.append((a0,b0))

    for _ in range(1,n):
        a,b = ints()
        flag = 0
        oria = a
        while stack:
            aa,bb = stack[-1]

            if b == bb:
                stack.pop()
                stack.append((a+aa,b))
                ans.append(max(ans[-1], a+aa))
                flag = 1
                break
            if aa >= oria:
                stack.append((oria,b))
                ans.append(ans[-1])
                flag = 1
                break

            else:
                a = min(oria - aa,a)
                stack.pop()
        if flag == 0:
            stack.append((oria,b))
            ans.append(oria)
    print(*ans)

# import sys
# for line in sys.stdin:
#     a,b = map(int, line.split())
#     print(a+b)

# python A.py < i.txt > o.txt

**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
4
1 0
1 1
1 0
1 1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Cosmic Rays
time_limit_ms: 2000
memory_limit_mb: 512
Description: Given an array of integers $s_1, s_2, \ldots, s_l$, every second, cosmic rays will cause all $s_i$ such that $i=1$ or $s_i\neq s_{i-1}$ to be deleted simultaneously, and the remaining parts will be concatenated together in order to form the new array $s_1, s_2, \ldots, s_{l'}$.

Define the strength of an array as the number of seconds it takes to become empty.

You are given an array of integers compressed in the form of $n$ pairs that describe the array left to right. Each pair $(a_i,b_i)$ represents $a_i$ copies of $b_i$, i.e. $\underbrace{b_i,b_i,\cdots,b_i}_{a_i\textrm{ times}}$.

For each $i=1,2,\dots,n$, please find the strength of the sequence described by the first $i$ pairs.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1\le t\le10^4$). The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($1\le n\le3\cdot10^5$) — the length of sequence $a$.

The next $n$ lines contain two integers each $a_i$, $b_i$ ($1\le a_i\le10^9,0\le b_i\le n$) — the pairs which describe the sequence.

It is guaranteed that the sum of all $n$ does not exceed $3\cdot10^5$.

It is guaranteed that for all $1\le i<n$, $b_i\neq b_{i+1}$ holds.

Output: For each test case, print one line containing $n$ integers — the answer for each prefix of pairs.

Examples:
input:
4
4
2 0
1 1
3 0
5 1
6
4 6
1 3
4 6
4 0
7 6
6 3
7
9 0
7 1
5 0
7 1
9 0
1 1
2 0
10
10 7
4 9
2 2
7 9
2 8
8 5
11 7
15 5
12 7
4 0
output:
2 2 4 5
4 4 7 7 10 10
9 9 9 9 9 9 10
10 10 10 10 10 10 12 15 15 15

Note: In the first test case, for the prefix of length $4$, the changes will be $[0,0,1,0,0,0,1,1,1,1,1]\rightarrow[0,0,0,1,1,1,1]\rightarrow[0,0,1,1,1]\rightarrow[0,1,1]\rightarrow[1]\rightarrow[]$, so the array becomes empty after $5$ seconds.

In the second test case, for the prefix of length $4$, the changes will be $[6,6,6,6,3,6,6,6,6,0,0,0,0]\rightarrow[6,6,6,6,6,6,0,0,0]\rightarrow[6,6,6,6,6,0,0]\rightarrow[6,6,6,6,0]\rightarrow[6,6,6]\rightarrow[6,6]\rightarrow[6]\rightarrow[]$, so the array becomes empty after $7$ seconds.

Buggy Code:
import java.util.*;

public final class e{
    public static void main(String[] args){
        Scanner scanner = new Scanner(System.in);
        int t = scanner.nextInt();
        scanner.nextLine();
        for(int i=0; i<t; i++){
            solve(scanner);
        }
        scanner.close();
    }

    static void solve(Scanner scanner){
        int n = scanner.nextInt();
        int pairs[][] = new int[n][2];
        for(int i=0; i<n; i++){
            pairs[i][0] = scanner.nextInt();
            pairs[i][1] = scanner.nextInt();
        }
        int s = pairs[0][0];
        Stack<int[]> stack = new Stack<>();
        for(int[] p: pairs){
            int last = 0;
            while(stack.isEmpty()==false && stack.peek()[1]!=p[1] && stack.peek()[0]<=p[0]){
                last = stack.pop()[0];
            }
            if(stack.isEmpty() || stack.peek()[1]!=p[1]){
                stack.push(p);
            }else{
                stack.peek()[0] += p[0]-last;
            }
            s = Math.max(s, stack.peek()[0]);
            System.out.print(s+" ");
        }
        System.out.println();
    }
}

**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
1 0
1 1
1 0
4
1 0
1 1
1 2
1 1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Cosmic Rays
time_limit_ms: 2000
memory_limit_mb: 512
Description: Given an array of integers $s_1, s_2, \ldots, s_l$, every second, cosmic rays will cause all $s_i$ such that $i=1$ or $s_i\neq s_{i-1}$ to be deleted simultaneously, and the remaining parts will be concatenated together in order to form the new array $s_1, s_2, \ldots, s_{l'}$.

Define the strength of an array as the number of seconds it takes to become empty.

You are given an array of integers compressed in the form of $n$ pairs that describe the array left to right. Each pair $(a_i,b_i)$ represents $a_i$ copies of $b_i$, i.e. $\underbrace{b_i,b_i,\cdots,b_i}_{a_i\textrm{ times}}$.

For each $i=1,2,\dots,n$, please find the strength of the sequence described by the first $i$ pairs.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1\le t\le10^4$). The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($1\le n\le3\cdot10^5$) — the length of sequence $a$.

The next $n$ lines contain two integers each $a_i$, $b_i$ ($1\le a_i\le10^9,0\le b_i\le n$) — the pairs which describe the sequence.

It is guaranteed that the sum of all $n$ does not exceed $3\cdot10^5$.

It is guaranteed that for all $1\le i<n$, $b_i\neq b_{i+1}$ holds.

Output: For each test case, print one line containing $n$ integers — the answer for each prefix of pairs.

Examples:
input:
4
4
2 0
1 1
3 0
5 1
6
4 6
1 3
4 6
4 0
7 6
6 3
7
9 0
7 1
5 0
7 1
9 0
1 1
2 0
10
10 7
4 9
2 2
7 9
2 8
8 5
11 7
15 5
12 7
4 0
output:
2 2 4 5
4 4 7 7 10 10
9 9 9 9 9 9 10
10 10 10 10 10 10 12 15 15 15

Note: In the first test case, for the prefix of length $4$, the changes will be $[0,0,1,0,0,0,1,1,1,1,1]\rightarrow[0,0,0,1,1,1,1]\rightarrow[0,0,1,1,1]\rightarrow[0,1,1]\rightarrow[1]\rightarrow[]$, so the array becomes empty after $5$ seconds.

In the second test case, for the prefix of length $4$, the changes will be $[6,6,6,6,3,6,6,6,6,0,0,0,0]\rightarrow[6,6,6,6,6,6,0,0,0]\rightarrow[6,6,6,6,6,0,0]\rightarrow[6,6,6,6,0]\rightarrow[6,6,6]\rightarrow[6,6]\rightarrow[6]\rightarrow[]$, so the array becomes empty after $7$ seconds.

Buggy Code:
import java.io.*;
import java.util.*;
public class E {
	public static PrintWriter out;
	static ArrayList<HashSet<Integer>>al;
	public static void main(String[] args)throws IOException{
        JS sc=new JS();
        out=new PrintWriter(System.out);
        int t=sc.nextInt();
        outer: while(t-->0) {
        	int n=sc.nextInt();
        	int a[]=new int[n];
        	int b[]=new int[n];
        	for(int i=0;i<n;i++) {
        		a[i]=sc.nextInt();
        		b[i]=sc.nextInt();
        	}
        	long ans=a[0];
        	Stack<Pair>stk=new Stack<>();
        	stk.add(new Pair(b[0], a[0]));
        	out.print(ans+" ");
        	for(int i=1;i<n;i++) {
        		if(stk.peek().x==b[i]) {
        			Pair cur=stk.pop();
        			cur.y+=a[i];
        		}else if(stk.peek().y<=a[i]) {
        			long max=0;
        			while(!stk.isEmpty()&&a[i]>=stk.peek().y&&stk.peek().x!=b[i]) {
        				max=Math.max(max, stk.pop().y);
        			}
        			if(stk.isEmpty()) {
        				stk.add(new Pair(b[i], a[i]));
        			}else if(stk.peek().x==b[i]) {
            			Pair cur=stk.pop();
            			cur.y+=(a[i]-max);
            			stk.add(cur);
        			}else {
        				stk.add(new Pair(b[i], a[i]));
        			}
        		}else if(stk.peek().y>a[i]) {
        			stk.add(new Pair(b[i], a[i]));
        		}
        		ans=Math.max(ans, stk.peek().y);
        		out.print(ans+" ");
        	}
        	out.println();
        }
		out.close();
	}
    public static long gcd(long a, long b) {
        if(a > b)
            a = (a+b)-(b=a);
        if(a == 0L)
            return b;
        return gcd(b%a, a);
    }
	static void add(HashMap<Character,Integer> A, char key){
		A.put(key,A.getOrDefault(key,0)+1);
	}
	static void remove(HashMap<Character,Integer>A, char key){
		if(!A.containsKey(key))return;
		A.put(key,A.get(key)-1);
		if(A.get(key)==0){
			A.remove(key);
		}
	}
	static class Pair{
		int x; long y;
		public Pair(int x, long y) {
			this.x=x;this.y=y;
		}
	}
	static void sort(long a[]) {
		ArrayList<Long>al=new ArrayList<>();
		for(int i=0;i<a.length;i++) {
			al.add(a[i]);
		}
		Collections.sort(al);
		for(int i=0;i<a.length;i++) {
			a[i]=al.get(i);
		}
	}
	static class JS {
	    public int BS = 1<<16;
	    public char NC = (char)0;
	    byte[] buf = new byte[BS];
	    int bId = 0, size = 0;
	    char c = NC;
	    double num = 1;
	    BufferedInputStream in;

	    public JS() {
	        in = new BufferedInputStream(System.in, BS);
	    }

	    public JS(String s) throws FileNotFoundException {
	        in = new BufferedInputStream(new FileInputStream(new File(s)), BS);
	    }

	    public char nextChar(){
	        while(bId==size) {
	            try {
	                size = in.read(buf);
	            }catch(Exception e) {
	                return NC;
	            }
	            if(size==-1)return NC;
	            bId=0;
	        }
	        return (char)buf[bId++];
	    }

	    public int nextInt() {
	        return (int)nextLong();
	    }

	    public long nextLong() {
	        num=1;
	        boolean neg = false;
	        if(c==NC)c=nextChar();
	        for(;(c<'0' || c>'9'); c = nextChar()) {
	            if(c=='-')neg=true;
	        }
	        long res = 0;
	        for(; c>='0' && c <='9'; c=nextChar()) {
	            res = (res<<3)+(res<<1)+c-'0';
	            num*=10;
	        }
	        return neg?-res:res;
	    }

	    public double nextDouble() {
	        double cur = nextLong();
	        return c!='.' ? cur:cur+nextLong()/num;
	    }

	    public String next() {
	        StringBuilder res = new StringBuilder();
	        while(c<=32)c=nextChar();
	        while(c>32) {
	            res.append(c);
	            c=nextChar();
	        }
	        return res.toString();
	    }

	    public String nextLine() {
	        StringBuilder res = new StringBuilder();
	        while(c<=32)c=nextChar();
	        while(c!='\n') {
	            res.append(c);
	            c=nextChar();
	        }
	        return res.toString();
	    }

	    public boolean hasNext() {
	        if(c>32)return true;
	        while(true) {
	            c=nextChar();
	            if(c==NC)return false;
	            else if(c>32)return true;
	        }
	    }
	}
}

**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 focus on edge cases and scenarios that can maximize the failure probability. The key aspects to consider are:

1. **Edge Cases:**
   - Arrays with a single element.
   - Arrays with repeated elements.
   - Arrays with alternating elements.
   - Arrays with large values for `a_i`.

2. **Scenarios to Maximize Failure Probability:**
   - Scenarios where the stack operations might lead to incorrect merging or deletion of elements.
   - Scenarios where the stack is empty or nearly empty, and new elements are added.
   - Scenarios where the elements are added in a way that the stack operations are complex and can lead to incorrect results.

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

```plaintext
3
1
1 0
2
1 0
1 1
3
1 0
1 1
1 0
```

### Explanation:

1. **Test Case 1:**
   - **Input:** `1 0`
   - **Output:** `1`
   - **Explanation:** The array is `[0]`, and it takes 1 second to become empty.

2. **Test Case 2:**
   - **Input:** `1 0 1 1`
   - **Output:** `1 2`
   - **Explanation:**
     - For the prefix `[0]`, it takes 1 second to become empty.
     - For the prefix `[0, 1]`, it takes 2 seconds to become empty.

3. **Test Case 3:**
   - **Input:** `1 0 1 1 1 0`
   - **Output:** `1 2 3`
   - **Explanation:**
     - For the prefix `[0]`, it takes 1 second to become empty.
     - For the prefix `[0, 1]`, it takes 2 seconds to become empty.
     - For the prefix `[0, 1, 0]`, it takes 3 seconds to become empty.

These test cases cover a range of scenarios, including single-element arrays, arrays with repeated elements, and arrays with alternating elements. They are designed to challenge the code and expose any potential bugs in the stack operations and element merging.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Cosmic Rays
time_limit_ms: 2000
memory_limit_mb: 512
Description: Given an array of integers $s_1, s_2, \ldots, s_l$, every second, cosmic rays will cause all $s_i$ such that $i=1$ or $s_i\neq s_{i-1}$ to be deleted simultaneously, and the remaining parts will be concatenated together in order to form the new array $s_1, s_2, \ldots, s_{l'}$.

Define the strength of an array as the number of seconds it takes to become empty.

You are given an array of integers compressed in the form of $n$ pairs that describe the array left to right. Each pair $(a_i,b_i)$ represents $a_i$ copies of $b_i$, i.e. $\underbrace{b_i,b_i,\cdots,b_i}_{a_i\textrm{ times}}$.

For each $i=1,2,\dots,n$, please find the strength of the sequence described by the first $i$ pairs.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1\le t\le10^4$). The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($1\le n\le3\cdot10^5$) — the length of sequence $a$.

The next $n$ lines contain two integers each $a_i$, $b_i$ ($1\le a_i\le10^9,0\le b_i\le n$) — the pairs which describe the sequence.

It is guaranteed that the sum of all $n$ does not exceed $3\cdot10^5$.

It is guaranteed that for all $1\le i<n$, $b_i\neq b_{i+1}$ holds.

Output: For each test case, print one line containing $n$ integers — the answer for each prefix of pairs.

Examples:
input:
4
4
2 0
1 1
3 0
5 1
6
4 6
1 3
4 6
4 0
7 6
6 3
7
9 0
7 1
5 0
7 1
9 0
1 1
2 0
10
10 7
4 9
2 2
7 9
2 8
8 5
11 7
15 5
12 7
4 0
output:
2 2 4 5
4 4 7 7 10 10
9 9 9 9 9 9 10
10 10 10 10 10 10 12 15 15 15

Note: In the first test case, for the prefix of length $4$, the changes will be $[0,0,1,0,0,0,1,1,1,1,1]\rightarrow[0,0,0,1,1,1,1]\rightarrow[0,0,1,1,1]\rightarrow[0,1,1]\rightarrow[1]\rightarrow[]$, so the array becomes empty after $5$ seconds.

In the second test case, for the prefix of length $4$, the changes will be $[6,6,6,6,3,6,6,6,6,0,0,0,0]\rightarrow[6,6,6,6,6,6,0,0,0]\rightarrow[6,6,6,6,6,0,0]\rightarrow[6,6,6,6,0]\rightarrow[6,6,6]\rightarrow[6,6]\rightarrow[6]\rightarrow[]$, so the array becomes empty after $7$ seconds.

Buggy Code:
#include <algorithm>
#include <array>
#include <cassert>
#include <cstdint>
#include <format>
#include <iostream>
#include <iterator>
#include <numeric>
#include <ostream>
#include <queue>
#include <ranges>
#include <string>
#include <vector>

#ifdef DYNAMIC_PIGEON
#include "algo/debug.h"
#else
#define debug(...) 114514
#define dprint(...) 114514
#endif

#define Dynamic_Pigeon 0

using i64 = std::int64_t;
using u64 = std::uint64_t;
using u32 = std::uint32_t;
using std::cin;
using std::cout;
using std::string;
using std::vector;

constexpr i64 MOD = 998244353;
constexpr int INF = 0x3f3f3f3f;
constexpr i64 INF_LONG_LONG = 0x3f3f3f3f3f3f3f3f;

#define all(a) a.begin(), a.end()
#define rall(a) a.rbegin(), a.rend()
#define print(s, ...) std::cout << std::format(s, ##__VA_ARGS__)
#define println(s, ...) std::cout << std::format(s, ##__VA_ARGS__) << "\n"

template <class T, class P>
concept CHANGE = requires(P b) { T(b); };

template <class T>
class SegmentTree {
public:
  struct node {
    T x, lazy;
    int l, r;
    int len() {
      return r - l + 1;
    }
  };

  SegmentTree(int n) : n(n), tree(4 * (n + 1)) {
    build(1, 1, n);
  }

  // 0 索引要空着
  template <class P>
    requires CHANGE<T, P>
  SegmentTree(const vector<P> &a) {
    n = a.size() - 1;
    tree.resize(4 * (n + 1));
    build(1, 1, n, a);
  }

  // 0 索引要空着
  template <class P>
    requires CHANGE<T, P>
  SegmentTree(const vector<P> &a, int n) : n(n), tree(4 * (n + 1)) {
    build(1, 1, n, a);
  }

  void update(int l, int r, T k) {
    update(1, l, r, 1, n, k);
  }
  void update(int pos, T k) {
    update(1, pos, pos, 1, n, k);
  }

  T query(int l, int r) {
    return query(1, l, r, 1, n);
  }
  T query(int pos) {
    return query(1, pos, pos, 1, n);
  }

private:
  int n;
  vector<node> tree;

  void pushup(int z) {
    tree[z].x = std::max(tree[z << 1].x, tree[z << 1 | 1].x);
  }
  void pushdown(int z) {
    // T lazy = tree[z].lazy;
    // if (bool(lazy)) {
    //   tree[z << 1].lazy += lazy;
    //   tree[z << 1].x += lazy * T(tree[z << 1].len());
    //   tree[z << 1 | 1].lazy += lazy;
    //   tree[z << 1 | 1].x += lazy * T(tree[z << 1 | 1].len());
    //   tree[z].lazy = 0;
    // }
  }

  template <class P>
    requires CHANGE<T, P>
  void build(int z, int l, int r, const vector<P> &a) {
    tree[z].l = l;
    tree[z].r = r;
    if (l == r) {
      tree[z].x = T(a[l]);
      return;
    }
    int mid = (l + r) >> 1;
    build(z << 1, l, mid, a);
    build(z << 1 | 1, mid + 1, r, a);
    pushup(z);
  }

  void build(int z, int l, int r) {
    tree[z].l = l;
    tree[z].r = r;
    if (l == r) {
      return;
    }
    int mid = (l + r) >> 1;
    build(z << 1, l, mid);
    build(z << 1 | 1, mid + 1, r);
    pushup(z);
  }
  void update(int z, int s, int e, int l, int r, T k) {
    if (l > e || r < s) {
      return;
    }
    if (l >= s && r <= e) {
      tree[z].x = k;
      return;
    }
    pushdown(z);
    int mid = (l + r) >> 1;
    update(z << 1, s, e, l, mid, k);
    update(z << 1 | 1, s, e, mid + 1, r, k);
    pushup(z);
  }
  T query(int z, int s, int e, int l, int r) {
    if (l > e || r < s) {
      return 0;
    }
    if (l >= s && r <= e) {
      return tree[z].x;
    }
    pushdown(z);
    int mid = (l + r) >> 1;
    return std::max(query(z << 1, s, e, l, mid),
                    query(z << 1 | 1, s, e, mid + 1, r));
  }
};

void tizzytyt_SuKi() {
  int n;
  cin >> n;
  vector<std::pair<int, int>> a(n + 1);
  for (int i = 1; i <= n; i++) {
    int l, r;
    cin >> l >> r;
    a[i] = {l, r};
  }

  vector<vector<int>> loc(n + 1);

  vector<int> pre(n + 1, 0);
  SegmentTree<i64> sum(n + 1);

  for (int i = 1; i <= n; i++) {
    int x = a[i].second;
    i64 t = a[i].first;
    while (!loc[x].empty() &&
           (sum.query(loc[x].back() + 1, i) >= t ||
            sum.query(loc[x].back() + 1, i) >= sum.query(loc[x].back()))) {
      loc[x].pop_back();
    }
    if (loc[x].empty()) {
      sum.update(i, t);
    } else {
      int pre = loc[x].back();
      loc[x].pop_back();
      i64 sm = sum.query(pre + 1, i);
      i64 cur = t - sm + sum.query(pre);
      sum.update(i, cur);
    }
    loc[x].push_back(i);
  }
  for (int i = 1; i <= n; i++) {
    cout << sum.query(1, i) << " ";
  }
  cout << "\n";
}
signed main() {
#ifdef DYNAMIC_PIGEON
  freopen("input.in", "r", stdin);
#else
  std::cin.tie(0)->sync_with_stdio(0);
#endif
  int t = 1;
  cin >> t;
  while (t--) {
    tizzytyt_SuKi();
  }
  return Dynamic_Pigeon;
}

**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 focus on edge cases that can cause the algorithm to fail. The key points to consider are:

1. **Edge Cases with Small Arrays**: Small arrays can help in quickly identifying issues with the algorithm's logic.
2. **Arrays with Repeated Values**: The algorithm should handle repeated values correctly, especially when they are followed by different values.
3. **Arrays with Large Values**: Large values for \(a_i\) can help in testing the limits of the algorithm.
4. **Arrays with Alternating Values**: Alternating values can expose issues in the logic for deleting elements.

Given these considerations, let's construct a test input that includes these edge cases:

1. **Small Array with Repeated Values**:
   - This will test the logic for handling repeated values and ensure that the algorithm correctly handles the deletion process.

2. **Array with Large Values**:
   - This will test the algorithm's ability to handle large numbers and ensure that it does not overflow or produce incorrect results.

3. **Array with Alternating Values**:
   - This will test the algorithm's ability to handle alternating values and ensure that it correctly processes the deletion steps.

Here is a small-scale test input that includes these edge cases:

```plaintext
3
3
1 0
1 1
1 0
4
1000000000 0
1 1
1000000000 0
1 1
5
1 0
1 1
1 0
1 1
1 0
```

This test input includes:
- A small array with repeated values.
- An array with large values.
- An array with alternating values.

These cases should help in identifying any issues with 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: Cosmic Rays
time_limit_ms: 2000
memory_limit_mb: 512
Description: Given an array of integers $s_1, s_2, \ldots, s_l$, every second, cosmic rays will cause all $s_i$ such that $i=1$ or $s_i\neq s_{i-1}$ to be deleted simultaneously, and the remaining parts will be concatenated together in order to form the new array $s_1, s_2, \ldots, s_{l'}$.

Define the strength of an array as the number of seconds it takes to become empty.

You are given an array of integers compressed in the form of $n$ pairs that describe the array left to right. Each pair $(a_i,b_i)$ represents $a_i$ copies of $b_i$, i.e. $\underbrace{b_i,b_i,\cdots,b_i}_{a_i\textrm{ times}}$.

For each $i=1,2,\dots,n$, please find the strength of the sequence described by the first $i$ pairs.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1\le t\le10^4$). The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($1\le n\le3\cdot10^5$) — the length of sequence $a$.

The next $n$ lines contain two integers each $a_i$, $b_i$ ($1\le a_i\le10^9,0\le b_i\le n$) — the pairs which describe the sequence.

It is guaranteed that the sum of all $n$ does not exceed $3\cdot10^5$.

It is guaranteed that for all $1\le i<n$, $b_i\neq b_{i+1}$ holds.

Output: For each test case, print one line containing $n$ integers — the answer for each prefix of pairs.

Examples:
input:
4
4
2 0
1 1
3 0
5 1
6
4 6
1 3
4 6
4 0
7 6
6 3
7
9 0
7 1
5 0
7 1
9 0
1 1
2 0
10
10 7
4 9
2 2
7 9
2 8
8 5
11 7
15 5
12 7
4 0
output:
2 2 4 5
4 4 7 7 10 10
9 9 9 9 9 9 10
10 10 10 10 10 10 12 15 15 15

Note: In the first test case, for the prefix of length $4$, the changes will be $[0,0,1,0,0,0,1,1,1,1,1]\rightarrow[0,0,0,1,1,1,1]\rightarrow[0,0,1,1,1]\rightarrow[0,1,1]\rightarrow[1]\rightarrow[]$, so the array becomes empty after $5$ seconds.

In the second test case, for the prefix of length $4$, the changes will be $[6,6,6,6,3,6,6,6,6,0,0,0,0]\rightarrow[6,6,6,6,6,6,0,0,0]\rightarrow[6,6,6,6,6,0,0]\rightarrow[6,6,6,6,0]\rightarrow[6,6,6]\rightarrow[6,6]\rightarrow[6]\rightarrow[]$, so the array becomes empty after $7$ seconds.

Buggy Code:
#include <bits/stdc++.h>
#define pii pair<ll, ll>
typedef long long ll;
using namespace std;

const int MAX = 300003;
pii P[MAX];

int main() {
	ios::sync_with_stdio(0); cin.tie(0);
	int _T;
	cin >> _T;
	while (_T--) {
		int N;
		cin >> N;
		for (int i = 1; i <= N; ++i) cin >> P[i].second >> P[i].first;

		vector<pii> ST;
		for (int i = 1; i <= N; ++i) {
			int erased = 0;
			while (!ST.empty() && ST.back().first != P[i].first && ST.back().second <= P[i].second) {
				erased = ST.back().second;
				ST.pop_back();
			}
			if (!ST.empty() && ST.back().first == P[i].first) {
				ST.back().second = ST.back().second + P[i].second - erased;
			}
			else ST.push_back(P[i]);

			cout << ST[0].second << ' ';
		}

		cout << '\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
3
1 0
1 1
1 0
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Cosmic Rays
time_limit_ms: 2000
memory_limit_mb: 512
Description: Given an array of integers $s_1, s_2, \ldots, s_l$, every second, cosmic rays will cause all $s_i$ such that $i=1$ or $s_i\neq s_{i-1}$ to be deleted simultaneously, and the remaining parts will be concatenated together in order to form the new array $s_1, s_2, \ldots, s_{l'}$.

Define the strength of an array as the number of seconds it takes to become empty.

You are given an array of integers compressed in the form of $n$ pairs that describe the array left to right. Each pair $(a_i,b_i)$ represents $a_i$ copies of $b_i$, i.e. $\underbrace{b_i,b_i,\cdots,b_i}_{a_i\textrm{ times}}$.

For each $i=1,2,\dots,n$, please find the strength of the sequence described by the first $i$ pairs.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1\le t\le10^4$). The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($1\le n\le3\cdot10^5$) — the length of sequence $a$.

The next $n$ lines contain two integers each $a_i$, $b_i$ ($1\le a_i\le10^9,0\le b_i\le n$) — the pairs which describe the sequence.

It is guaranteed that the sum of all $n$ does not exceed $3\cdot10^5$.

It is guaranteed that for all $1\le i<n$, $b_i\neq b_{i+1}$ holds.

Output: For each test case, print one line containing $n$ integers — the answer for each prefix of pairs.

Examples:
input:
4
4
2 0
1 1
3 0
5 1
6
4 6
1 3
4 6
4 0
7 6
6 3
7
9 0
7 1
5 0
7 1
9 0
1 1
2 0
10
10 7
4 9
2 2
7 9
2 8
8 5
11 7
15 5
12 7
4 0
output:
2 2 4 5
4 4 7 7 10 10
9 9 9 9 9 9 10
10 10 10 10 10 10 12 15 15 15

Note: In the first test case, for the prefix of length $4$, the changes will be $[0,0,1,0,0,0,1,1,1,1,1]\rightarrow[0,0,0,1,1,1,1]\rightarrow[0,0,1,1,1]\rightarrow[0,1,1]\rightarrow[1]\rightarrow[]$, so the array becomes empty after $5$ seconds.

In the second test case, for the prefix of length $4$, the changes will be $[6,6,6,6,3,6,6,6,6,0,0,0,0]\rightarrow[6,6,6,6,6,6,0,0,0]\rightarrow[6,6,6,6,6,0,0]\rightarrow[6,6,6,6,0]\rightarrow[6,6,6]\rightarrow[6,6]\rightarrow[6]\rightarrow[]$, so the array becomes empty after $7$ seconds.

Buggy Code:
def get_strength(pairs):
    # Initialize the strength counter
    strength = 0

    # Reconstruct the array from the pairs
    arr = []
    for count, value in pairs:
        arr += [value] * count

    # Simulate the deletion process until the array is empty
    while len(arr) > 0:
        # Increment the strength counter
        strength += 1

        # Create a new list to store the updated array
        new_arr = []

        # Iterate over the elements of the original array
        for i in range(len(arr)):
            # If it's the first element or its value differs from the previous one, delete it
            if i == 0 or arr[i] != arr[i - 1]:
                continue
            # Otherwise, append it to the new array
            else:
                new_arr.append(arr[i])

        # Update the original array with the new one
        arr = new_arr

    return strength

# Read the number of test cases
t = int(input())
for _ in range(t):
    # Read the number of pairs for each test case
    n = int(input())
    # Read the pairs themselves
    pairs = [tuple(map(int, input().split())) for _ in range(n)]

    # Calculate and output the strength for each prefix
    result = []
    for i in range(1, n + 1):
        result.append(get_strength(pairs[:i]))
    print(*result)

**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
3
1 0
1 1
1 0
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Cosmic Rays
time_limit_ms: 2000
memory_limit_mb: 512
Description: Given an array of integers $s_1, s_2, \ldots, s_l$, every second, cosmic rays will cause all $s_i$ such that $i=1$ or $s_i\neq s_{i-1}$ to be deleted simultaneously, and the remaining parts will be concatenated together in order to form the new array $s_1, s_2, \ldots, s_{l'}$.

Define the strength of an array as the number of seconds it takes to become empty.

You are given an array of integers compressed in the form of $n$ pairs that describe the array left to right. Each pair $(a_i,b_i)$ represents $a_i$ copies of $b_i$, i.e. $\underbrace{b_i,b_i,\cdots,b_i}_{a_i\textrm{ times}}$.

For each $i=1,2,\dots,n$, please find the strength of the sequence described by the first $i$ pairs.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1\le t\le10^4$). The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($1\le n\le3\cdot10^5$) — the length of sequence $a$.

The next $n$ lines contain two integers each $a_i$, $b_i$ ($1\le a_i\le10^9,0\le b_i\le n$) — the pairs which describe the sequence.

It is guaranteed that the sum of all $n$ does not exceed $3\cdot10^5$.

It is guaranteed that for all $1\le i<n$, $b_i\neq b_{i+1}$ holds.

Output: For each test case, print one line containing $n$ integers — the answer for each prefix of pairs.

Examples:
input:
4
4
2 0
1 1
3 0
5 1
6
4 6
1 3
4 6
4 0
7 6
6 3
7
9 0
7 1
5 0
7 1
9 0
1 1
2 0
10
10 7
4 9
2 2
7 9
2 8
8 5
11 7
15 5
12 7
4 0
output:
2 2 4 5
4 4 7 7 10 10
9 9 9 9 9 9 10
10 10 10 10 10 10 12 15 15 15

Note: In the first test case, for the prefix of length $4$, the changes will be $[0,0,1,0,0,0,1,1,1,1,1]\rightarrow[0,0,0,1,1,1,1]\rightarrow[0,0,1,1,1]\rightarrow[0,1,1]\rightarrow[1]\rightarrow[]$, so the array becomes empty after $5$ seconds.

In the second test case, for the prefix of length $4$, the changes will be $[6,6,6,6,3,6,6,6,6,0,0,0,0]\rightarrow[6,6,6,6,6,6,0,0,0]\rightarrow[6,6,6,6,6,0,0]\rightarrow[6,6,6,6,0]\rightarrow[6,6,6]\rightarrow[6,6]\rightarrow[6]\rightarrow[]$, so the array becomes empty after $7$ seconds.

Buggy Code:
from typing import List, Tuple
import sys

def calculate_strength(arr: List[int]) -> int:
    if not arr:
        return 0

    seconds = 0
    while len(arr) > 0:
        new_arr = []
        prev = None
        for num in arr:
            if prev is None or num != prev:
                prev = num
            else:
                new_arr.append(num)
        arr = new_arr
        seconds += 1
    return seconds

def main():
    input_lines = sys.stdin.read().strip().split('\n')
    index = 0
    t = int(input_lines[index])
    index += 1
    results = []

    for _ in range(t):
        n = int(input_lines[index])
        index += 1
        pairs = [tuple(map(int, input_lines[index + i].split())) for i in range(n)]
        index += n

        strengths = []
        current_array = []

        for a_i, b_i in pairs:
            current_array.extend([b_i] * a_i)
            strengths.append(calculate_strength(current_array))

        results.append(' '.join(map(str, strengths)))

    for result in results:
        print(result)

if __name__ == "__main__":
    main()

**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
4
1 0
1 1
1 0
1 1
6
1 0
1 1
1 0
1 1
1 0
1 1
7
1 0
1 1
1 0
1 1
1 0
1 1
1 0
```