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 <algorithm>
#include <iostream>
#include <cstring>
#include <cstdio>
#include <cmath>
using namespace std;

const int MaxN = 500;
const int MaxM = MaxN * MaxN;

int n;
int pos[MaxN + 1];

int mark[MaxM + 1];
int used[MaxN + 1];

int seq[MaxM + 1];

inline bool construct()
{
	for (int i = 1; i <= n * n; ++i)
	{
		if (mark[i] > 0)
		{
			int k = mark[i];
			if (++used[k] != k)
				return false;
			seq[i] = k;
			continue;
		}

		int cur = 0;
		for (int k = 1; k <= n; ++k)
		{
			if (used[k] >= k - 1 || i > pos[k])
				continue;
			if (cur == 0 || pos[k] < pos[cur])
				cur = k;
		}

		for (int k = 1; !cur && k <= n; ++k)
			if (used[k] < n && i > pos[k])
				cur = k;

		if (!cur)
			return false;

		seq[i] = cur, ++used[cur];
	}

	return true;
}

int main()
{
	cin >> n;
	for (int i = 1; i <= n; ++i)
	{
		cin >> pos[i];
		mark[pos[i]] = i;
	}

	if (!construct())
		puts("No");
	else
	{
		puts("Yes");
		for (int i = 1; i <= n * n; ++i)
			printf("%d ", seq[i]);
	}

	return 0;
}