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 ll;
typedef double db;
typedef pair<int,int> pii;
#define eprintf(...) fprintf(stderr,__VA_ARGS__)

const int N=505;

int n,l,r;
int a[N*N];
pii t[N];

int main(){
	scanf("%d",&n);
	for(int i=1;i<=n;i++) t[i].second=i,scanf("%d",&t[i].first);
	sort(t+1,t+n+1);
	for(int i=1;i<=n;i++) a[t[i].first]=t[i].second;
	bool ok=true;
	l=1;
	for(int i=1;i<=n;i++){
		int cnt=t[i].second-1;
		while(cnt--){
			while(a[l]) l++;
			if(l>t[i].first) ok=false;
			a[l++]=t[i].second;
		}
	}
	r=n*n;
	for(int i=n;i>=1;i--){
		int cnt=n-t[i].second;
		while(cnt--){
			while(a[r]) r--;
			if(r<t[i].first) ok=false;
			a[r--]=t[i].second;
		}
	}
	if(!ok)
		puts("No");
	else{
		puts("Yes");
		for(int i=1;i<=n*n;i++) printf("%d%c",a[i]," \n"[i==n*n]);
	}
}