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;
typedef long long lint;
typedef long double llf;
typedef pair<int, int> pi;

set<int> s;

int n, a[505];
int r[250005];
vector<int> v[505];
pi b[505];

int main(){
	cin >> n;
	for(int i=1; i<=n; i++){
		cin >> a[i];
		b[i] = pi(a[i], i);
	}
	sort(b+1, b+n+1);
	for(int i=1; i<=n*n; i++){
		s.insert(i);
	}
	for(int i=1; i<=n; i++){
		for(int j=1; j<b[i].second; j++){
			if(*s.begin() >= b[i].first){
				puts("No");
				return 0;
			}
			r[*s.begin()] = b[i].second;
			s.erase(s.begin());
		}
		s.erase(b[i].first);
		r[b[i].first] = b[i].second;
	}
	for(int i=1; i<=n; i++){
		for(int j=b[i].second+1; j<=n; j++){
			r[*s.begin()] = b[i].second;
			s.erase(s.begin());
		}
	}
	for(int i=1; i<=n*n; i++){
		v[r[i]].push_back(i);
	}
	for(int i=1; i<=n; i++){
		if(v[i][i-1] != a[i]){
			puts("No");
			return 0;
		}
	}
	puts("Yes");
	for(int i=1; i<=n*n; i++) printf("%d ",r[i]);
}

