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>

#define For(i, j, k) for(int i = j; i <= k; i++)
#define Forr(i, j, k) for(int i = j; i >= k; i--)

using namespace std;

const int N = 510;

struct Number{
	int o, x;

	bool operator < (const Number& A) const{
		return x < A.x;
	}
}A[N];

int n;
int P[N * N];

int main(){
	
	scanf("%d", &n);
	For(i, 1, n) scanf("%d", &A[i].x), A[i].o = i;
	sort(A + 1, A + n + 1);

	int pos = 1;
	For(i, 1, n){
		P[A[i].x] = A[i].o;
		For(j, 1, A[i].o - 1){
			while(P[pos]) ++pos;
			P[pos] = A[i].o;
		}

		if(pos > A[i].x){
			puts("No");
			return 0;
		}
	}

	pos = n * n;
	Forr(i, n, 1){
		For(j, A[i].o + 1, n){
			while(P[pos]) --pos;
			P[pos] = A[i].o;
		}

		if(pos < A[i].x){
			puts("No");
			return 0;
		}
	}

	puts("Yes");
	For(i, 1, n * n) printf("%d%c", P[i], i == n * n ? '\n' : ' ');
	return 0;
}