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 N 250010
#define halt return puts("No"),0
inline int read() {
    int x=0;
    char ch=getchar();
    while (!isdigit(ch)) ch=getchar();
    while (isdigit(ch)) x=x*10+ch-'0',ch=getchar();
    return x;
}
int p[N],c[N],a[N];
int main() {
    int n=read();
    for (int i=1;i<=n;i++) p[i]=read();
    for (int i=1;i<=n*n;i++) {
        for (int j=1;j<=n;j++) if (p[j]==i) {
            if ((++c[j])!=j) halt; a[i]=j; break;
        }
        if (a[i]) continue;
        int Min=0;
        for (int j=1;j<=n;j++)
            if (c[j]<j-1 && (!Min || p[j]<p[Min])) Min=j;
        if (Min) {c[a[i]=Min]++;continue;}
        for (int j=1;j<=n;j++) 
            if (c[j]<n && p[j]<i) {c[a[i]=j]++;break;}
        if (!a[i]) halt;
    }
    puts("Yes");
    for (int i=1;i<=n*n;i++) printf("%d ",a[i]);
    return 0;
}