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>
int arr[505], sz_l[505], sz_r[505], ans[300005];
int main()
{
	// freopen("AGC008-D.in", "r", stdin);
	int n;
	scanf("%d", &n);
	for (int i = 0; i < n; i++)
	{
		scanf("%d", arr + i);
		arr[i]--;
	}
	for (int i = 0; i < n * n; i++)
	{
		ans[i] = -1;
		for (int j = 0; j < n && -1 == ans[i]; j++)
		{
			if (arr[j] == i)
				ans[i] = j;
		}
		int pos = -1;
		for (int j = 0; j < n && -1 == ans[i]; j++)
		{
			if (i < arr[j] && sz_l[j] < j && (pos == -1 || arr[j] < arr[pos]))
				pos = j;
		}
		if (~pos)
		{
			ans[i] = pos;
			sz_l[pos]++;
		}
		for (int j = 0; j < n && -1 == ans[i]; j++)
		{
			if (arr[j] < i && sz_r[j] < n - j - 1)
			{
				ans[i] = j;
				sz_r[j]++;
			}
		}
		if (ans[i] == -1)
		{
			puts("No");
			return 0;
		}
	}
	puts("Yes");
	for (int i = 0; i < n * n; i++)
		printf("%d%c", ans[i] + 1, " \n"[i + 1 == n * n]);
	return 0;
}
