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<bits/stdc++.h>
using namespace std;
#define ll long long
#define f(i, x, n) for(int i = x; i < (int)n; ++i)

int n, an[500 * 500 + 2], p = 1;
pair<int, int> x[500];

void no(){
	printf("No\n");
	exit(0);
}

void fill(int id, int f, int l, int r){
	while (an[p])++p;
	while (f && p >= l && p <= r){
		if (!an[p])an[p] = id, --f;
		++p;
	}
	if (f)no();
}

int main(){
	scanf("%d", &n);
	f(i, 0, n)scanf("%d", &x[i].first), x[i].second = i + 1;
	sort(x, x + n);
	f(i, 0, n){
		fill(x[i].second, x[i].second - 1, 1, x[i].first - 1);
		an[x[i].first] = x[i].second;
	}
	f(i, 0, n)fill(x[i].second, n - x[i].second, x[i].first + 1, n * n);
	printf("Yes\n%d", an[1]);
	int nn = n * n;
	f(i, 2, nn + 1)printf(" %d", an[i]);
	printf("\n");
}
