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>

const int N = 540, N2 = 254000;

int n, n2;
int pos[N], a[N2];
int o[N], remain[N];

int main() {
	int i, j;
	scanf("%d", &n), n2 = n * n;
	for (i = 1; i <= n; ++i) scanf("%d", pos + i), a[pos[i]] = i, o[i - 1] = i;
	std::sort(o, o + n, [] (const int x, const int y) {return pos[x] < pos[y];});
	for (i = 1; i <= n; ++i) remain[i] = i - 1;
	j = !remain[*o];
	for (i = 1; i <= n2 && j < n; ++i)
		if (!a[i]) for (--remain[ a[i] = o[j] ]; j < n && !remain[o[j]]; ++j);
		else if (remain[a[i]]) break;
	if (j < n) return puts("No"), 0;
	for (i = 1; i <= n; ++i) remain[i] = n - i;
	j = (remain[o[n - 1]] ? n - 1 : n - 2);
	for (i = n2; i && j >= 0; --i)
		if (!a[i]) for (--remain[ a[i] = o[j] ]; j >= 0 && !remain[o[j]]; --j);
		else if (remain[a[i]]) break;
	if (j >= 0) return puts("No"), 0;
	puts("Yes");
	for (i = 1; i <= n2; ++i) printf("%d%c", a[i], i == n2 ? 10 : 32);
	return 0;
}