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 <cstdio>
#include <iostream>
#include <vector>
using namespace std;

const int MaxN = 505;
int a[MaxN*MaxN], x[MaxN*MaxN];

int main(){
	int n; scanf("%d",&n);
	for(int i=1; i<=n; ++i){
		int r; scanf("%d",&r);
		x[r] = i; // 直接存
	}
	bool ok = true;
	for(int i=1; i<=n*n; ++i){
		if(x[i] == 0) continue;
		int cnt = a[i] = x[i];
		for(int j=1; j<i&&cnt>1; ++j)
			if(a[j] == 0)
				a[j] = x[i], -- cnt;
		if(cnt > 1) ok = false;
	}
	for(int i=n*n; i>=1; --i){
		if(x[i] == 0) continue;
		int cnt = n-x[i];
		for(int j=n*n; j>i&&cnt>0; --j)
			if(a[j] == 0)
				a[j] = x[i], -- cnt;
		if(cnt > 0) ok = false;
	}
	if(ok){
		printf("Yes\n");
		for(int i=1; i<=n*n; ++i)
			printf("%d ",a[i]);
	}
	else printf("No\n");
	return 0;
}