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;

typedef long double ld;
typedef long long ll;

const int N = 505;
const int M = N * N;

int n, m;
int cnt[N];
int x[N];

void read() {
  cin >> n;
  m = n * n;

  for (int i = 0; i < n; ++i) {
    cin >> x[i];
    --x[i];
  }
}

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

void kill() {
  vector<int> ans;

  for (int j = 0; j < m; ++j) {
    int y = -1;
    for (int i = 0; i < n; ++i) {
      if (j == x[i]) {
        y = i;
        if (cnt[i] != i)
          no();
        break;
      }

      if (j > x[i]) {
        if (y == -1 && cnt[i] < n)
          y = i;
        continue;
      }

      if ((y == -1 || x[y] < j || x[i] < x[y]) && cnt[i] < i) {
        y = i;
      }
    }

    if (y == -1)
      no();

    cnt[y]++;
    ans.push_back(y);
  }

  cout << "Yes\n";
  for (int x : ans)
    cout << x + 1 << " ";
  cout << endl;
}

int main() {
#ifdef LOCAL
  assert(freopen("d.in", "r", stdin));
#endif

  ios_base::sync_with_stdio(false);

  read();
  kill();
}

