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>
#include <queue>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>
typedef long long ll;
#define rep(i, a, b) for (int i = a; i <= b; ++ i)
#define per(i, a, b) for (int i = a; i >= b; -- i)
const int maxn = 250010;
using namespace std;
int a[maxn], b[maxn], n;
priority_queue <pair<int, int> > p; 
int main() {
	scanf("%d", &n); rep(i, 1, n) {
		scanf("%d", &b[i]), a[b[i]] = i;
		rep(j, 1, i - 1) p.push(make_pair(- (b[i] - 1), i));
	}
	rep(i, 1, n * n) {
		if (a[i]) {
			rep(j, a[i] + 1, n) p.push(make_pair(- (n * n + 1), a[i]));
		} else {
			if (p.empty() || p.top().first > - i) {
				printf("No");
				return 0;
			}
			a[i] = p.top().second;
			p.pop();
		}
	}
	printf("Yes\n");
	rep(i, 1, n * n) printf("%d ", a[i]);
	return 0;
}