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 <cstdio>
#include <vector>
#include <utility>
#include <algorithm>
using namespace std;

vector<pair<int, int> > p;
int r[250010];

int main(){
  int N; scanf("%d", &N);
  for(int i = 1; i <= N; i++){
    int x; scanf("%d", &x); r[x] = i;
    p.push_back(make_pair(x, i));
  }

  sort(p.begin(), p.end());

  bool v = true;

  int i_ = 1;
  for(int i = 0; i < N; i++){
    int x = p[i].first, y = p[i].second;

    for(int c = 0; c < y - 1; c++){
      while(r[i_] != 0) i_++;
      r[i_++] = y;
    }

    if(i_ > x) v = false;
  }

  i_ = N * N;
  for(int i = N - 1; i >= 0; i--){
    int x = p[i].first, y = p[i].second;

    for(int c = 0; c < N - y; c++){
      while(r[i_] != 0) i_--;
      r[i_--] = y;
    }

    if(i_ < x) v = false;
  }

  puts(v ? "Yes" : "No");
  if(v){
    for(int i = 1; i <= N * N; i++) printf("%d ", r[i]);
    printf("\n");
  }
  return 0;
}