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;

int a[501*501];
int x[501];
int n;

pair<int, int> v[501];

void no() {
  cout << "No\n";
  exit(0);
}

int main() {
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> x[i];
    v[i] = {x[i], i};
  }
  sort(v + 1, v + n + 1);
  int last = 1;
  for (int i = 1; i <= n; i++) {
    a[v[i].first] = v[i].second;
    for (int j = 0; j < v[i].second - 1; j++) {
      while (a[last] != 0) last++;
      if (last >= v[i].first) no();
      a[last] = v[i].second;
    }
  }
  
  last = n*n;
  for (int i = n; i >= 1; --i) {
    for (int j = 0; j < n - v[i].second; j++) {
      while (a[last] != 0) last--;
      if (last <= v[i].first) { no(); }
      a[last] = v[i].second;
    }
  }
  
  cout << "Yes\n";
  for (int i = 1; i <= n * n; i++) {
    cout << a[i] << ' ';
  }

  return 0;
}