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>
#define rep(i,k,n) for(int i=k;i<=n;i++)
#define rep2(i,k,n) for(int i=k;i>=n;i--)
using namespace std;
const int N=505;
const int M=3e5+7;
int a[M],n,m,vis[N];
queue<int> Q;
int main(){
	scanf("%d",&n);
	rep(i,1,n){
		int x;scanf("%d",&x);
		a[x]=i;
	}
	m=n*n;
	rep(i,1,m){
		if(a[i]){
			rep(j,1,a[i]-1){
				if(Q.empty()){
					puts("No");
					return 0;
				}
				a[Q.front()]=a[i];
				Q.pop();
			}
		}
		else Q.push(i);
	}
	while(!Q.empty())Q.pop();
	rep2(i,m,1){
		if(a[i] && !vis[a[i]]){
			vis[a[i]]=1;
			rep(j,1,n-a[i]){
				if(Q.empty()){
					puts("No");
					return 0;
				}
				a[Q.front()]=a[i];
				Q.pop();
			}
		}
		else Q.push(i);
	}
	puts("Yes");
	rep(i,1,m)printf("%d ",a[i]);
}