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<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<vector>
using namespace std;
const int N=510;
struct Data{
	int id,val;
	friend bool operator < (Data x,Data y){return x.val<y.val;}
}a[N];
vector<int> L[N],R[N];
int ans[N*N];
int n,m;
bool solve(){
	for (int now=1;now<=n*n;++now){
		for (int i=1;i<=n&&!ans[now];++i){
			if (a[i].val==now) ans[now]=a[i].id;
		}
		for (int i=1;i<=n&&!ans[now];++i){
			if (L[a[i].id].size()<a[i].id-1&&now<a[i].val)
				ans[now]=a[i].id,L[a[i].id].push_back(now);
		}
		for (int i=1;i<=n&&!ans[now];++i){
			if (R[a[i].id].size()<n-a[i].id&&now>a[i].val)
				ans[now]=a[i].id,R[a[i].id].push_back(now);
		}
		if (!ans[now]) return false;
	}
	return true;
}

int main(){
#ifndef ONLINE_JUDGE
	//freopen("a.in","r",stdin);
#endif
	scanf("%d",&n);
	for (int i=1;i<=n;++i) scanf("%d",&a[i].val),a[i].id=i;
	sort(a+1,a+1+n);
	if (!solve()) printf("No\n");
	else{
		printf("Yes\n");
		for (int i=1;i<=n*n;++i) printf("%d ",ans[i]); printf("\n");
	}
}