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<stdio.h>
#include<vector>
#include<algorithm>
using namespace std;
typedef pair<int, int>pii;
int ans[1000000];
int main()
{
	int num;
	scanf("%d", &num);
	vector<pii>v;
	for (int i = 0; i < num; i++)
	{
		int z;
		scanf("%d", &z);
		v.push_back(make_pair(z, i + 1));
		ans[z] = i + 1;
	}
	sort(v.begin(), v.end());
	int pt = 1;
	for (int i = 0; i < v.size(); i++)
	{
		for (int j = 0; j < v[i].second - 1; j++)
		{
			for (;;)
			{
				if (ans[pt] == 0)
				{
					if (pt >= v[i].first)goto l01;
					ans[pt] = v[i].second;
					pt++;
					break;
				}
				pt++;
			}
		}
	}
	reverse(v.begin(), v.end());
	pt = num * num;
	for (int i = 0; i < v.size(); i++)
	{
		for (int j = 0; j < num - v[i].second; j++)
		{
			for (;;)
			{
				if (ans[pt] == 0)
				{
					if (pt <= v[i].first)goto l01;
					ans[pt] = v[i].second;
					pt--;
					break;
				}
				pt--;
			}
		}
	}
	printf("Yes\n");
	for (int i = 0; i < num*num; i++)printf("%d ", ans[i + 1]); printf("\n");
	if (false)
	{
	l01:;
		printf("No\n");
	}
}