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 <vector>
#include <algorithm>
#include <cstdlib>
using namespace std;

const int MAXN = 500;
int n, x[MAXN+1];
int a[MAXN*MAXN+1];

vector<pair<int,int>> q;

void fk(){
	puts("No");
	exit(0);
}

int main(){
	scanf("%d", &n);
	for(int i=1; i<=n; i++){
		scanf("%d", &x[i]);
		a[x[i]]=i;
		q.push_back(make_pair(x[i], i));
	}

	sort(q.begin(), q.end());
	int index=1;
	for(auto &p:q){
		int place=p.first, num=p.second;
		for(int i=1; i<num; i++){
			while(index<n*n && a[index]) index++;
			if(a[index] || index > place) fk(); ///
			a[index]=num; index++;
		}
	}

	for(int i=1; i<=n; i++){
        for(int j=1, k=x[i]; j<=n-i; j++){
			while(k<n*n && a[k]) k++;
			if(a[k]) fk(); ///
			a[k]=i;
        }
	}

	puts("Yes");
	for(int i=1; i<=n*n; i++)
		printf("%d ", a[i]);
	return 0;
}
