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
//by Sshwy
#include<bits/stdc++.h>
using namespace std;
#define pb push_back
#define FOR(i,a,b) for(int i=(a);i<=(b);++i)
#define ROF(i,a,b) for(int i=(a);i>=(b);--i)

namespace RA{
    int rnd(int p){return 1ll*rand()*rand()%p;}
    int rnd(int L,int R){return rnd(R-L+1)+L;}
}
const int N=505;
int n;
int a[N];
int b[N*N];
int c[N],ok[N]; // counter of usage
vector<int> v;
int main(){
    scanf("%d",&n);
    FOR(i,1,n)scanf("%d",&a[i]),b[a[i]]=i;
    FOR(i,1,n*n)if(b[i])v.pb(b[i]);
    FOR(i,1,n*n){
        if(b[i]){
            if(c[b[i]]+1==b[i]) c[b[i]]++,ok[b[i]]=1;
            else return puts("No"),0;
        }else {
            bool flag=0;
            for(int j:v)if(!ok[j] && c[j]<j-1){ ++c[j],b[i]=j,flag=1;break; }
            if(flag==0)for(int j:v)if(ok[j] && c[j]<n){ ++c[j],b[i]=j,flag=1;break; }
            if(flag==0)return puts("No"),0;
        }
    }
    puts("Yes");
    FOR(i,1,n*n)printf("%d%c",b[i]," \n"[i==n*n]);
    return 0;
}
