Phoenix is trying to take a photo of his n friends with labels 1, 2, ..., n who are lined up in a row in a special order. But before he can take the photo, his friends get distracted by a duck and mess up their order.

Now, Phoenix must restore the order but he doesn't remember completely! He only remembers that the i-th friend from the left had a label between a_i and b_i inclusive. Does there exist a unique way to order his friends based of his memory? 

Input

The first line contains one integer n (1 ≤ n ≤ 2⋅10^5) — the number of friends.

The i-th of the next n lines contain two integers a_i and b_i (1 ≤ a_i ≤ b_i ≤ n) — Phoenix's memory of the i-th position from the left.

It is guaranteed that Phoenix's memory is valid so there is at least one valid ordering.

Output

If Phoenix can reorder his friends in a unique order, print YES followed by n integers — the i-th integer should be the label of the i-th friend from the left.

Otherwise, print NO. Then, print any two distinct valid orderings on the following two lines. If are multiple solutions, print any.

Examples

Input


4
4 4
1 3
2 4
3 4


Output


YES
4 1 2 3 


Input


4
1 3
2 4
3 4
2 3


Output


NO
1 3 4 2 
1 2 4 3 
4
3 4
1 2
3 4
1 2
 4
2 4
1 3
4 4
1 1
 10
6 6
1 2
2 2
6 7
8 8
3 5
5 5
4 4
8 9
10 10
 4
1 4
1 4
2 4
1 2
 8
3 4
7 8
2 6
1 3
6 7
5 8
4 6
8 8
 7
7 7
2 3
2 3
5 7
5 6
4 4
1 2
 15
2 2
10 11
2 5
6 7
7 12
9 9
2 3
5 14
1 3
10 10
15 15
12 12
5 6
6 6
14 15
 5
1 3
4 4
5 5
1 4
1 5
 4
1 4
2 3
2 3
4 4
 4
1 4
1 4
1 2
1 4
 5
1 3
1 3
1 5
3 4
1 5
 3
1 3
2 2
1 3
 5
4 4
1 4
3 4
1 3
3 5
 3
1 3
1 2
2 3
 4
2 4
1 3
2 3
1 2
 5
4 5
4 5
1 3
1 2
2 4
 6
2 6
4 6
6 6
2 2
3 3
1 3
 3
2 3
1 2
1 3
 6
4 6
1 3
4 6
4 4
1 4
1 2
 7
3 6
6 6
5 7
2 4
2 3
1 2
3 5
 8
4 6
8 8
1 4
4 5
6 8
1 2
3 6
1 4
 6
2 4
2 6
4 5
2 5
6 6
1 4
 8
4 4
7 7
3 5
5 7
1 8
4 5
1 5
2 2
 4
3 4
1 3
2 3
1 2
 10
9 9
3 4
1 1
5 5
7 7
3 7
6 10
3 3
10 10
1 3
 4
1 4
4 4
2 3
1 1
 1
1 1
 8
1 5
6 8
1 3
4 8
1 1
3 3
6 6
8 8
 5
3 3
4 4
2 2
1 1
5 5
 4
1 1
2 3
2 3
1 4
 4
3 4
1 4
2 3
1 2
 2
1 2
1 2
 8
3 3
3 5
2 3
1 1
5 7
7 7
5 5
3 8
 5
2 2
3 3
1 2
1 5
4 4
 6
1 3
2 4
3 4
2 3
5 5
6 6
 6
1 4
3 6
6 6
3 5
1 5
2 5
 5
4 4
2 3
4 5
1 4
2 4
 3
1 2
1 2
1 3
 4
1 4
2 2
3 3
1 4
 5
2 5
1 1
2 3
3 4
5 5
 4
2 4
1 4
3 4
1 3
 6
1 1
4 6
3 6
1 3
3 4
3 6
 5
2 3
3 4
4 5
1 1
2 3
 5
2 3
1 5
2 4
3 5
4 4
 20
12 14
13 13
18 18
3 3
7 8
11 11
10 13
2 2
9 10
3 5
7 8
15 17
14 15
9 9
20 20
15 16
6 6
19 19
1 1
5 5
 5
1 5
2 2
3 3
4 4
1 5
 7
5 7
1 5
1 2
1 5
5 6
3 5
7 7

NO
3 1 4 2
3 2 4 1
 NO
3 2 4 1 
2 3 4 1  YES
6 1 2 7 8 3 5 4 9 10  NO
2 3 4 1 
1 3 4 2 
 YES
3 7 2 1 6 5 4 8  NO
7 2 3 6 5 4 1 
7 3 2 6 5 4 1 
 YES
2 11 4 7 8 9 3 13 1 10 15 12 5 6 14  NO
1 4 5 2 3 
2 4 5 1 3  NO
1 2 3 4 
1 3 2 4 
 NO
2 3 1 4 
1 3 2 4 
 NO
1 2 4 3 5 
2 1 4 3 5 
 NO
1 2 3 
3 2 1 
 NO
4 2 3 1 5 
4 1 3 2 5  NO
2 1 3
1 2 3
 NO
4 2 3 1
4 1 3 2
 NO
4 5 2 1 3
4 5 1 2 3
 NO
4 5 6 2 3 1 
5 4 6 2 3 1 
 NO
2 1 3 
3 1 2 
 NO
5 2 6 4 3 1 
5 1 6 4 3 2 
 NO
5 6 7 3 2 1 4 
5 6 7 2 3 1 4  NO
5 8 2 4 7 1 6 3
5 8 1 4 7 2 6 3
 NO
2 5 4 3 6 1 
3 5 4 2 6 1  YES
4 7 3 6 8 5 1 2  NO
4 2 3 1
4 1 3 2
 YES
9 4 1 5 7 6 8 3 10 2  NO
3 4 2 1 
2 4 3 1  YES
1  NO
4 7 2 5 1 3 6 8 
5 7 2 4 1 3 6 8  YES
3 4 2 1 5  NO
1 2 3 4 
1 3 2 4 
 NO
3 4 2 1 
4 3 2 1 
 NO
1 2 
2 1 
 YES
3 4 2 1 6 7 5 8  YES
2 3 1 5 4  NO
1 3 4 2 5 6
1 2 4 3 5 6
 NO
1 5 6 3 2 4 
2 5 6 3 1 4 
 NO
4 2 5 1 3 
4 3 5 1 2  NO
1 2 3 
2 1 3 
 NO
1 2 3 4 
4 2 3 1 
 NO
4 1 2 3 5 
3 1 2 4 5 
 NO
2 3 4 1
2 1 4 3
 NO
1 4 5 2 3 6 
1 5 4 2 3 6 
 NO
2 4 5 1 3 
3 4 5 1 2 
 NO
2 1 3 5 4 
3 1 2 5 4  NO
14 13 18 3 7 11 12 2 10 4 8 17 15 9 20 16 6 19 1 5 
14 13 18 3 8 11 12 2 10 4 7 17 15 9 20 16 6 19 1 5 
 NO
1 2 3 4 5 
5 2 3 4 1 
 NO
6 2 1 3 5 4 7
6 1 2 3 5 4 7

#pragma comment(linker, "/stack:200000000")
//#pragma GCC optimize("Ofast")
//#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
//#pragma GCC target ("avx2")
//#pragma GCC optimization ("unroll-loops")
#pragma GCC optimize ("O3")
#pragma GCC target ("sse4")

#include <bits/stdc++.h>
#include <ext/pb_ds/tree_policy.hpp>
#include <ext/pb_ds/assoc_container.hpp>

using namespace std;
using namespace __gnu_pbds;
template <class T> using Tree = tree<T, null_type, less<T>, rb_tree_tag,tree_order_statistics_node_update>;

const long double PI = 4*atan(1);

typedef long long ll;
typedef long double ld;

typedef pair<int, int> pi;
typedef pair<ll, ll> pl;
typedef pair<ld, ld> pd;
typedef complex<ld> cd;

typedef vector<int> vi;
typedef vector<ll> vl;
typedef vector<ld> vd;
typedef vector<pi> vpi;
typedef vector<pl> vpl;
typedef vector<cd> vcd;

#define f0r(i,a) for(int i=0;i<a;i++)
#define f1r(i,a,b) for(int i=a;i<b;i++)
#define trav(a, x) for (auto& a : x)
#define FOR(i,a,b) for (int i = (a); i < (b); ++i)
#define F0R(i,a) FOR(i,0,a)
#define ROF(i,a,b) for (int i = (b)-1; i >= (a); --i)
#define R0F(i,a) ROF(i,0,a)

#define mp make_pair
#define pb push_back
#define eb emplace_back
#define f first
#define s second
#define lb lower_bound
#define ub upper_bound

#define sz(x) (int)(x).size()
#define all(x) x.begin(), x.end()
#define rall(x) rbegin(x), rend(x)
#define resz resize
#define ins insert

template<typename A, typename B> ostream& operator<<(ostream &cout, pair<A, B> const &p) { return cout << "(" << p.f << ", " << p.s << ")"; }
template<typename A> ostream& operator<<(ostream &cout, vector<A> const &v) {
  cout << "["; for(int i = 0; i < v.size(); i++) {if (i) cout << ", "; cout << v[i];} return cout << "]";
}

template<class T> bool ckmin(T& a, const T& b) {
	return a > b ? a = b, 1 : 0; }
template<class T> bool ckmax(T& a, const T& b) {
	return a < b ? a = b, 1 : 0; }

namespace input {
	template<class T> void re(complex<T>& x);
	template<class T1, class T2> void re(pair<T1,T2>& p);
	template<class T> void re(vector<T>& a);
	template<class T, size_t SZ> void re(array<T,SZ>& a);

	template<class T> void re(T& x) { cin >> x; }
	void re(double& x) { string t; re(t); x = stod(t); }
	void re(ld& x) { string t; re(t); x = stold(t); }
	template<class T, class... Ts> void re(T& t, Ts&... ts) {
		re(t); re(ts...);
	}

	template<class T> void re(complex<T>& x) { T a,b; re(a,b); x = cd(a,b); }
	template<class T1, class T2> void re(pair<T1,T2>& p) { re(p.f,p.s); }
	template<class T> void re(vector<T>& a) { F0R(i,sz(a)) re(a[i]); }
	template<class T, size_t SZ> void re(array<T,SZ>& a) { F0R(i,SZ) re(a[i]); }
}

using namespace input;

namespace output {
	void pr(int x) { cout << x; }
	void pr(long x) { cout << x; }
	void pr(ll x) { cout << x; }
	void pr(unsigned x) { cout << x; }
	void pr(unsigned long x) { cout << x; }
	void pr(unsigned long long x) { cout << x; }
	void pr(float x) { cout << x; }
	void pr(double x) { cout << x; }
	void pr(ld x) { cout << x; }
	void pr(char x) { cout << x; }
	void pr(const char* x) { cout << x; }
	void pr(const string& x) { cout << x; }
	void pr(bool x) { pr(x ? "true" : "false"); }
	template<class T> void pr(const complex<T>& x) { cout << x; }

	template<class T1, class T2> void pr(const pair<T1,T2>& x);
	template<class T> void pr(const T& x);

	template<class T, class... Ts> void pr(const T& t, const Ts&... ts) {
		pr(t); pr(ts...);
	}
	template<class T1, class T2> void pr(const pair<T1,T2>& x) {
		pr("{",x.f,", ",x.s,"}");
	}
	template<class T> void pr(const T& x) {
		pr("{"); // const iterator needed for vector<bool>
		bool fst = 1; for (const auto& a: x) pr(!fst?", ":"",a), fst = 0;
		pr("}");
	}

	void ps() { pr("\n"); } // print w/ spaces
	template<class T, class... Ts> void ps(const T& t, const Ts&... ts) {
		pr(t); if (sizeof...(ts)) pr(" "); ps(ts...);
	}

	void pc() { pr("]\n"); } // debug w/ commas
	template<class T, class... Ts> void pc(const T& t, const Ts&... ts) {
		pr(t); if (sizeof...(ts)) pr(", "); pc(ts...);
	}
	#define dbg(x...) pr("[",#x,"] = ["), pc(x);
}

using namespace output;

namespace io {
	void setIn(string s) { freopen(s.c_str(),"r",stdin); }
	void setOut(string s) { freopen(s.c_str(),"w",stdout); }
	void setIO(string s = "") {
		cin.sync_with_stdio(0); cin.tie(0); // fast I/O
		// cin.exceptions(cin.failbit); // ex. throws exception when you try to read letter into int
		if (sz(s)) { setIn(s+".in"), setOut(s+".out"); } // for USACO
	}
}

using namespace io;
mt19937 rng((uint32_t)chrono::steady_clock::now().time_since_epoch().count());

const ll MOD = 1e9 + 7;

typedef decay<decltype(MOD)>::type T;
struct mi {
	T val;
	explicit operator T() const { return val; }
	mi() { val = 0; }
	mi(const ll& v) {
		val = (-MOD <= v && v <= MOD) ? v : v % MOD;
		if (val < 0) val += MOD;
	}
	// friend ostream& operator<<(ostream& os, const mi& a) {
		// return os << a.val; }
	friend void pr(const mi& a) { pr(a.val); }
	friend void re(mi& a) { ll x; re(x); a = mi(x); }

	friend bool operator==(const mi& a, const mi& b) {
		return a.val == b.val; }
	friend bool operator!=(const mi& a, const mi& b) {
		return !(a == b); }
	friend bool operator<(const mi& a, const mi& b) {
		return a.val < b.val; }

	mi operator-() const { return mi(-val); }
	mi& operator+=(const mi& m) {
		if ((val += m.val) >= MOD) val -= MOD;
		return *this; }
	mi& operator-=(const mi& m) {
		if ((val -= m.val) < 0) val += MOD;
		return *this; }
	mi& operator*=(const mi& m) {
		val = (ll)val*m.val%MOD; return *this; }
	friend mi pow(mi a, ll p) {
		mi ans = 1; assert(p >= 0);
		for (; p; p /= 2, a *= a) if (p&1) ans *= a;
		return ans;
	}
	friend mi inv(const mi& a) {
		assert(a != 0); return pow(a,MOD-2); }
	mi& operator/=(const mi& m) { return (*this) *= inv(m); }

	friend mi operator+(mi a, const mi& b) { return a += b; }
	friend mi operator-(mi a, const mi& b) { return a -= b; }
	friend mi operator*(mi a, const mi& b) { return a *= b; }
	friend mi operator/(mi a, const mi& b) { return a /= b; }
};

typedef pair<mi,mi> pmi;
typedef vector<mi> vmi;
typedef vector<pmi> vpmi;
const int MAX = 2e5 + 5;
int a[MAX];
int b[MAX];
vector<pair<pi, int>> v;
int ans[MAX];
int n;
void print(vi use){
    f0r(i, n){
        int id = v[i].s;
        ans[id] = use[i];
    }
    f0r(i, n) pr(ans[i]+1,  " ");
    ps();
}
int main(){
    re(n);
    f0r(i, n) re(a[i], b[i]);

    f0r(i, n){
        a[i]--; b[i]--;
        v.eb(mp(mp(a[i], b[i]), i));
    }
    sort(all(v));
    vi v1;
    vi v2;
    f0r(i, n) v1.eb(i);
    f0r(j, n) v2.eb(j);
    f0r(i, n){
        if(v[i].f.f != i){
            ps("NO");

            swap(v2[i], v2[i+1]);
            print(v1);
            print(v2);
            return 0;
        }
    }
    ps("YES");
    print(v1);
    return 0;
}
