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

#include <bits/stdc++.h>
//#pragma GCC optimize("unroll-loops")
//#pragma GCC optimize("-O3")
//#pragma GCC optimize("Ofast")
//#pragma GCC optimize("fast-math")
//#pragma GCC optimize("no-stack-protector")
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#include <ext/rope>
#define sz(x) int(x.size())
#define all(x) x.begin(),x.end()
#define PB push_back
#define MP make_pair
#define pii pair<int, int>
#define pll pair<ll, ll>
#define pil pair<int, ll>
#define pli pair<ll, int>
#define pdd pair<ld, ld>
#define ft first
#define sd second
using namespace std;
using namespace __gnu_cxx;
using namespace __gnu_pbds;
typedef long long ll;
typedef long double ld;
template<class T>
using ordered_set = tree<T,null_type,less<T>,rb_tree_tag,tree_order_statistics_node_update>;
const int N = 200100;
const int BIG = int(5e6);
const int oo = 2e9;
const ll OO = 1e18;
const int md = 998244353;
const ll PW = 100003;
set<array<int, 3> > st;
int nm[N], l[N], r[N], n, a[N], b[N];

mt19937 rnd(chrono::system_clock().now().time_since_epoch().count());

bool cmp(int _x, int _y){
    return l[_x] < l[_y];
}

bool cmp1(int _x, int _y){
    return r[_x] > r[_y];
}

int main() {
#ifdef _LOCAL
    freopen("in.txt","r",stdin); //freopen("output.txt","w",stdout);
#else
//    freopen("mining.in","r",stdin); freopen("mining.out","w",stdout);
    ios_base::sync_with_stdio(0); cin.tie(0);
#endif

    cin >> n;

    for (int i = 0; i < n; i++){
        cin >> l[i] >> r[i];
        nm[i] = i;
    }

    sort(nm, nm + n, cmp);

    st.clear();

    for (int i = 1, ptr = 0; i <= n; i++){
        while (ptr < n && l[nm[ptr]] <= i){
            st.insert({r[nm[ptr]], 0, nm[ptr]});
            ptr++;
        }

        int pos = (*st.begin())[2];

        a[pos] = i;

        st.erase(st.begin());
    }

    //second part

    bool ok = 0;

    st.clear();

    for (int i = 1, ptr = 0; i <= n; i++){
        while (ptr < n && l[nm[ptr]] <= i){
            st.insert({r[nm[ptr]], 0, nm[ptr]});
            ptr++;
        }

        int pos = (*st.begin())[2];

        array<int, 3> cur = (*st.begin());

        b[pos] = i;

        if (b[pos] != a[pos])
            ok = 1;

        st.erase(st.begin());

        if (sz(st) && !ok){
            int np = (*st.begin())[2];

            if (cur[0] >= i + 1){
                b[pos] = i + 1;
                b[np] = i;
                st.erase(st.begin());
                ok = 1;
                i++;
            }
        }
    }

    if (!ok){
        sort(nm, nm + n, cmp1);

        st.clear();

        for (int i = n, ptr = 0; i > 0; i--){
            while (ptr < n && r[nm[ptr]] >= i){
                st.insert({-l[nm[ptr]], 0, -nm[ptr]});
                ptr++;
            }

            int pos = (*st.begin())[2];

            array<int, 3> cur = (*st.begin());

            b[-pos] = i;

            if (b[-pos] != a[-pos])
                ok = 1;

            st.erase(st.begin());

            if (sz(st) && !ok){
                int np = (*st.begin())[2];

                if (-cur[0] <= i - 1){
                    b[-pos] = i - 1;
                    b[-np] = i;
                    st.erase(st.begin());
                    ok = 1;
                    i--;
                }
            }
        }
    }

    if (ok){
        cout << "NO\n";
        for (int i = 0; i < n; i++)
            cout << a[i] << " "; cout << '\n';
        for (int i = 0; i < n; i++)
            cout << b[i] << " "; cout << '\n';
    } else {
        cout << "YES\n";
        for (int i = 0; i < n; i++)
            cout << a[i] << " "; cout << '\n';
    }

    return 0;
}
