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>
using namespace std;
int f[250009];
int x[509];
int S[509];
int main(){
    int n;
    cin >> n;
    vector<pair<int,int> > X;
    for(int i = 1; i <= n; i++){
        cin >> x[i];
        f[x[i]] = i;
        X.push_back({x[i],i});

    }
    sort(X.begin(), X.end());
    vector<int> v;
    for(int i = 0; i < n; i++){
        for(int j = 0; j < X[i].second - 1; j++)
            v.push_back(X[i].second),S[X[i].second]++;
    }
    int k = 0;
    for(int i = 1; i <= n*n; i++){
        if(f[i]){
            if(S[f[i]]){
                cout<<"No"<<endl;
                return 0;
            }
            for(int j = 0; j < n - f[i]; j++)
                v.push_back(f[i]),
                S[f[i]]++;
            continue;

        }
        if(k >= v.size()) {
            cout << "No"<<endl;
            return 0;
        }
        f[i] = v[k];
        S[f[i]] --;
        k ++;
    }
    cout<< "Yes"<<endl;
    for(int i = 1; i <= n * n; i++){
        cout << f[i] <<" ";
    }
    cout<<endl;


}
