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<bits/stdc++.h>

const int N = 510;
typedef std::pair <int, int> pii;

int x[N], a[N * N];
pii p[N];

void no(){
	puts("No");
	exit(0);
}

int main(){
	int n;
	scanf("%d", &n);
	for (int i = 1; i <= n; ++ i){
		scanf("%d", &x[i]);
		a[x[i]] = i;
		p[i] = {x[i], i};
	}
	std::sort(p + 1, p + n + 1);
	std::set <int> set;
	for (int i = 1; i <= n * n; ++ i){
		if (!a[i]) set.insert(i);
	}
	for (int i = 1; i <= n; ++ i){
		for (int j = 1; j < p[i].second; ++ j){
			int sit = *(set.begin());
			if (sit > p[i].first) no();
			a[sit] = p[i].second;
			set.erase(sit);
		}
	}
	for (int i = n; i >= 1; -- i){
		for (int j = p[i].second + 1; j <= n; ++ j){
			int sit = *(-- set.end());
			if (sit < p[i].first) no();
			a[sit] = p[i].second;
			set.erase(sit);
		}
	}
	puts("Yes");
	for (int i = 1; i <= n * n; ++ i){
		printf("%d%c", a[i], " \n"[i == n * n]);
	}
	return 0;
}