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;
#define lor(a,b,c) for(register int a=b;a<=c;++a)
#define ror(a,b,c) for(register int a=c;a>=b;--a)

const int N=505;

int n,x[N],y[N*N],z,cnt[N]; bool vis[N];
struct data {int pos,id;};
inline bool operator < (data a,data b) {return a.pos>b.pos;}
priority_queue <data> line;

int main(){
	scanf("%d",&n); lor(i,1,n) scanf("%d",&x[i]);
	lor(i,1,n) y[x[i]]=i; lor(i,1,n) lor(j,1,i-1) line.push((data){x[i],i});
	lor(i,1,n*n){
		z=i;
		if(y[i]){
			if(cnt[y[i]]!=y[i]-1) {return puts("No"),0;}
			else vis[y[i]]=true;
		}
		else if(!line.empty()){
			data tmp=line.top(); line.pop(); y[i]=tmp.id; cnt[tmp.id]++;
		}
		else {--z; break;}
	}
	lor(i,1,n) cnt[i]=n-i;
	lor(i,z+1,n*n){
		if(y[i]) vis[y[i]]=true;
		else{
			int ch=0;
			lor(j,1,n) if(cnt[j]&&vis[j]) {ch=j; cnt[ch]--; y[i]=ch; break;}
			if(!ch) return puts("No"),0;
		}
	}
	puts("Yes");
	lor(i,1,n*n) printf("%d ",y[i]); puts("");

	return 0;
}
