You are given an integer sequence x of length N. Determine if there exists an integer sequence a that satisfies all of the following conditions, and if it exists, construct an instance of a.

* a is N^2 in length, containing N copies of each of the integers 1, 2, ..., N.
* For each 1 ≤ i ≤ N, the i-th occurrence of the integer i from the left in a is the x_i-th element of a from the left.

Constraints

* 1 ≤ N ≤ 500
* 1 ≤ x_i ≤ N^2
* All x_i are distinct.

Input

The input is given from Standard Input in the following format:


N
x_1 x_2 ... x_N


Output

If there does not exist an integer sequence a that satisfies all the conditions, print `No`. If there does exist such an sequence a, print `Yes` in the first line, then print an instance of a in the second line, with spaces inbetween.

Examples

Input

3
1 5 9


Output

Yes
1 1 1 2 2 2 3 3 3


Input

2
4 1


Output

No
#include <bits/stdc++.h>
typedef long long ll;
using namespace std;
const ll N=1e5+5;
pair< ll , ll > a[505];
ll ans[505*505];
int main(){
    ios_base:: sync_with_stdio(false);
    cin.tie(NULL); cout.tie(NULL);
    ll n;cin>>n;
    for(ll i=1;i<=n;i++){
        cin>>a[i].first;
        a[i].second=i;
        ans[a[i].first]=a[i].second;
    }
    sort(a+1,a+n+1);
    ll now=1;
    for(ll i=1;i<=n;i++){
        for(ll j=1;j<a[i].second;j++){
            while(ans[now] && now<=n*n)now++;
            if(now>n*n)return cout<<"No"  , 0;
            ans[now]=a[i].second;
        }
        if(now > a[i].first )return cout<<"No" , 0;
    }
    now=n*n;
    for(ll i=n;i>=1;i--){
        for(ll j=a[i].second+1;j<=n;j++){
            while(ans[now] && now>=0 )now--;
            ans[now]=a[i].second;
        }
        if(now<a[i].first)return cout<<"No" , 0;
    }

    for(ll i=1;i<=n*n;i++)if(!ans[i])return cout<<"No" , 0;

    cout<<"Yes\n";
    for(ll i=1;i<=n*n;i++)cout<<ans[i]<<" ";
}

