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>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>

using namespace std;
using namespace __gnu_pbds;

typedef tree<int, null_type, less<int>, rb_tree_tag,tree_order_statistics_node_update> ordered_set;

#define scan(x) do{while((x=getchar())<'0'); for(x-='0'; '0'<=(_=getchar()); x=(x<<3)+(x<<1)+_-'0');}while(0)
char _;
#define complete_unique(a) a.erase(unique(a.begin(),a.end()),a.end())
#define all(a) a.begin(),a.end()
#define println printf("\n");
#define readln(x) getline(cin,x);
#define pb push_back
#define endl "\n"
#define INT_INF 0x3f3f3f3f
#define LL_INF 0x3f3f3f3f3f3f3f3f
#define EPS 1e-9
#define MOD 1000000007
#define MOD2 1494318097
#define SEED 131
#define mp make_pair
#define fastio cin.tie(0); cin.sync_with_stdio(0);

#define MAXN 200005

typedef unsigned long long ull;
typedef long long ll;
typedef long double ld;
typedef pair<int,int> pii;
typedef pair<double,double> pdd;
typedef pair<ll,ll> pll;
typedef pair<int,pii> triple;
typedef int8_t byte;

const ld PI = (ld)4*atanl(1);

mt19937 g1(chrono::steady_clock::now().time_since_epoch().count());

int randint(int a, int b){return uniform_int_distribution<int>(a, b)(g1);}
ll randlong(ll a,ll b){return uniform_int_distribution<long long>(a, b)(g1);}

ll gcd(ll a, ll b){return b == 0 ? a : gcd(b, a % b);}
ll lcm(ll a, ll b){return a*b/gcd(a,b);}
ll fpow(ll  b, ll exp, ll mod){if(exp == 0) return 1;ll t = fpow(b,exp/2,mod);if(exp&1) return t*t%mod*b%mod;return t*t%mod;}
ll divmod(ll i, ll j, ll mod){i%=mod,j%=mod;return i*fpow(j,mod-2,mod)%mod;}

struct node{
    int l,r;
    vector<int> s;
};

struct MergeSortTree{
    node seg[4*MAXN];
    inline void merge(int rt){
        vector<int> f = seg[rt<<1].s, s = seg[rt<<1|1].s;
        std::merge(all(f),all(s),back_inserter(seg[rt].s));
    }
    void build(int l, int r, int rt, int (&arr)[MAXN]){
        seg[rt].l = l, seg[rt].r = r, seg[rt].s.clear();
        if(l == r){
            seg[rt].s = {arr[l]};
            return;
        }
        int mid = (l+r)/2;
        build(l,mid,rt<<1,arr);
        build(mid+1,r,rt<<1|1,arr);
        merge(rt);
    }
    int query(int l, int r, int rt, int val, int s){
        if(l > r) return 0;
        if(seg[rt].l == l && seg[rt].r == r){
            if(s == 0) return lower_bound(all(seg[rt].s),val)-seg[rt].s.begin();
            return seg[rt].s.end()-upper_bound(all(seg[rt].s),val);
        }
        int mid = (seg[rt].l+seg[rt].r)/2;
        if(r <= mid) return query(l,r,rt<<1,val,s);
        else if(l > mid) return query(l,r,rt<<1|1,val,s);
        return query(l,mid,rt<<1,val,s)+query(mid+1,r,rt<<1|1,val,s);
    }
};

int len,res[MAXN],pos[MAXN],tmp[MAXN],tmp2[MAXN],used[MAXN];
MergeSortTree lft,rht;
pair<pii,int> arr[MAXN];
set<int> rem;

int main(){
    fastio; cin >> len;
    for(int i=1; i<=len; i++){
        cin >> arr[i].first.first >> arr[i].first.second;
        arr[i].second = i;
    }
    sort(arr+1,arr+len+1);
    for(int i=1; i<=len; i++)
        rem.insert(i);
    for(int i=1; i<=len; i++){
        assert(rem.size());
        int nxt = *rem.lower_bound(arr[i].first.first);
        res[arr[i].second] = nxt;
        rem.erase(nxt);
        pos[nxt] = i;
        used[i] = nxt;
    }
    for(int i=1; i<=len; i++){
        tmp[i] = arr[pos[i]].first.first;
        tmp2[i] = arr[pos[i]].first.second;
//        cout << tmp[i] << " ";
    }
//    for(int i=1; i<=len; i++) cout << tmp[i] << " "; cout << endl;
//    for(int i=1; i<=len; i++) cout << tmp2[i] << " "; cout << endl;
//    cout << endl;
    lft.build(1,len,1,tmp);
    rht.build(1,len,1,tmp);
    for(int i=1; i<=len; i++){
        pii range = arr[pos[i]].first;
        int am = rht.query(range.first,range.second,1,i,0);
        am+=lft.query(range.first,range.second,1,i,1);
        if(am != range.second-range.first){
            cout << "NO" << endl;
            for(int i=1; i<=len; i++)
                cout << res[i] << " ";
            cout << endl;
            int idx = -1;
            for(int k=1; k<=len; k++){
                if(i == k) continue;
                if(arr[k].first.first <= i && arr[k].first.second >= i && range.first <= used[k] && range.second >= used[k]){
                    idx = k;
                    break;
                }
            }
            assert(idx!=-1);
            int ind1 = arr[i].second, ind2 = arr[idx].second;
            swap(res[ind1],res[ind2]);
            for(int i=1; i<=len; i++)
                cout << res[i] << " ";
            return 0;
        }
    }
    cout << "YES" << endl;
    for(int i=1; i<=len; i++)
        cout << res[i] << " ";
}