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 <iostream>
#include <cmath>
#include <cstdio>
#include <cstring>
#include <algorithm>
#define rep(x, a, b) for(int x = (a); x <= (b); ++x)
#define rop(x, a, b) for(int x = (a); x < (b); ++x)
#define per(x, a, b) for(int x = (a); x >= (b); --x)
using namespace std;
typedef long long LL;
typedef double DB;
int x[505], a[2500005];
int n, vs[2500005];
int main() {
	scanf("%d", &n);
	rep(i, 1, n) scanf("%d", &x[i]), a[x[i]] = i, vs[x[i]] = 1;
	int p = 0;
	rep(i, 1, n * n) {
		if(vs[i] == 0) continue;
		int y = a[i] - 1;
		while(y--) {
			++p;
			while(a[p]) ++p;
			a[p] = a[i];
		}
		if(p >= i) {puts("No"); return 0;}
	}
	rep(i, 1, n * n) {
		if(vs[i] == 0) continue;
		int y = n - a[i];
		
		while(y--) {
			++p;
			while(a[p]) ++p;
			if(p <= i) {puts("No"); return 0;}
			a[p] = a[i];
		}
	}
	puts("Yes");
	rep(i, 1, n * n) printf("%d ", a[i]);
	return 0;
}
