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 <cstdio>
#include <queue>
#define N 505

using namespace std;

int n;
int a[N * N], rem[N];

struct node {
	int pos, val;
}q[N];
inline bool cmp_node(node lhs, node rhs) { return lhs.pos < rhs.pos; }

queue<int> que;

int main() {
	cin >> n;
	for(int i = 1; i <= n; ++i) {
		rem[i] = i - 1;
		cin >> q[i].pos, a[q[i].pos] = i;
		q[i].val = i;
	}
	
	sort(q + 1, q + 1 + n, cmp_node);
	for(int i = 1; i <= n; ++i)
		for(int j = 1; j < q[i].val; ++j)
			que.push(q[i].val);
	
	for(int i = 1; i <= n * n; ++i) {
		if(a[i]) {
			if(rem[a[i]]) {
				puts("No");
				return 0;
			} else {
				for(int j = 1; j <= n - a[i]; ++j)
					que.push(a[i]);
			}
		} else {
			if(!que.empty()) {
				a[i] = que.front();
				--rem[a[i]];
				que.pop();
			} else {
				puts("No");
				return 0;
			}
		}
	}
	
	puts("Yes");
	for(int i = 1; i <= n * n; ++i)
		printf("%d%c", a[i], i == n * n ? '\n' : ' ');
	return 0;
}