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>

using namespace std;

int n,i,j,ans[1000005],cnt[505];
pair <int, int> a[505];
queue <int> ok;

int main(){
	scanf("%d", &n);
	for(i = 1; i <= n; i++){
		scanf("%d", &a[i].first);
		a[i].second = i;
		ans[a[i].first] = i;
		cnt[i]++;
	}
	sort(a + 1, a + 1 + n);
	
	int p = 1;
	
	for(i = 1; i <= n * n; i++){
		
		while(p <= n && cnt[a[p].second] == a[p].second)
		p++;
		
		
		if(p <= n && i == a[p].first) return 0 * printf("No\n");
		
		if(p <= n && ans[i] == 0){
			int id = a[p].second;
			ans[i] = id;
			cnt[id]++;
		}
	}
	
	p = 1;
	
	for(i = 1; i <= n * n; i++){
		int id = a[p].second;
		if(a[p].first == i){
			ok.push(id);
			p++;
		}
		
		if(ans[i] != 0) continue;
		
		while(ok.size() && cnt[ok.front()] == n)
		ok.pop();
		
		if(ok.empty())
		return 0 * printf("No\n");
		
		ans[i] = ok.front();
		cnt[ok.front()]++;
	}
	
	printf("Yes\n");
	for(i = 1; i <= n * n; i++){
		if(i == 1)
		printf("%d", ans[i]);
		else
		printf(" %d", ans[i]);
	}
	printf("\n");
}
