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 N 510
using namespace std;
int x[N],a[N*N],id[N],od[N];
bool cmp(int i,int j){
	return x[i]<x[j];
}
int main(){
	int n,nxt=1;
	queue<int> must,pad;
	bool gg=false;
	scanf("%d",&n);
	for(int i=1;i<=n;i++){
		scanf("%d",&x[i]);
		id[i]=i;
	}
	sort(id+1,id+n+1,cmp);
	for(int i=1;i<=n;i++) od[id[i]]=i;
	for(int i=1;i<=n;i++){
		for(int j=0;j<id[i]-1;j++){
			must.push(id[i]);
		}
	}
	for(int i=1;i<=n*n;i++){
		if(i==x[id[nxt]]){
			if(!must.empty()&&id[nxt]>1&&od[must.front()]<=nxt){
				gg=true;
				break;
			}
			a[i]=id[nxt];
			for(int j=0;j<n-id[nxt];j++){
				pad.push(id[nxt]);
			}
			nxt++;
		}
		else if(!must.empty()){
			a[i]=must.front();
			must.pop();
		}
		else if(!pad.empty()){
			a[i]=pad.front();
			pad.pop();
		}
		else{
			gg=true;
			break;
		}
	}
	if(gg) puts("No");
	else{
		puts("Yes");
		for(int i=1;i<=n*n;i++){
			if(i>1) putchar(' ');
			printf("%d",a[i]);
		}
		puts("");
	}
	return 0;
}