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,n)for(int i=0;i<(n);i++)
using namespace std;
typedef pair<int, int>P;

P x[500];
int a[250000];
int main() {
	int n; scanf("%d", &n);
	rep(i, n) {
		scanf("%d", &x[i].first);
		a[--x[i].first] = i + 1;
		x[i].second = i + 1;
	}
	sort(x, x + n);
	int s = 0;
	rep(i, n) {
		rep(j, x[i].second - 1) {
			while (s < x[i].first&&a[s] != 0)s++;
			if (s == x[i].first) {
				puts("No"); return 0;
			}
			a[s] = x[i].second;
		}
	}
	s = 0;
	rep(i, n) {
		s = max(s, x[i].first + 1);
		rep(j, n - x[i].second) {
			while (s < n*n&&a[s] != 0)s++;
			if (s == n*n) {
				puts("No"); return 0;
			}
			a[s] = x[i].second;
		}
	}
	puts("Yes");
	rep(i, n*n) {
		if (i)printf(" ");
		printf("%d", a[i]);
	}
	puts("");
}