Problem: There are n piles of pebbles on the table, the i-th pile contains ai pebbles. Your task is to paint each pebble using one of the k given colors so that for each color c and any two piles i and j the difference between the number of pebbles of color c in pile i and number of pebbles of color c in pile j is at most one.

In other words, let's say that bi, c is the number of pebbles of color c in the i-th pile. Then for any 1 ≤ c ≤ k, 1 ≤ i, j ≤ n the following condition must be satisfied |bi, c - bj, c| ≤ 1. It isn't necessary to use all k colors: if color c hasn't been used in pile i, then bi, c is considered to be zero.

Input

The first line of the input contains positive integers n and k (1 ≤ n, k ≤ 100), separated by a space — the number of piles and the number of colors respectively.

The second line contains n positive integers a1, a2, ..., an (1 ≤ ai ≤ 100) denoting number of pebbles in each of the piles.

Output

If there is no way to paint the pebbles satisfying the given condition, output "NO" (without quotes) .

Otherwise in the first line output "YES" (without quotes). Then n lines should follow, the i-th of them should contain ai space-separated integers. j-th (1 ≤ j ≤ ai) of these integers should be equal to the color of the j-th pebble in the i-th pile. If there are several possible answers, you may output any of them.

Examples

Input

4 4
1 2 3 4


Output

YES
1
1 4
1 2 4
1 2 3 4


Input

5 2
3 2 4 1 3


Output

NO


Input

5 4
3 2 4 3 5


Output

YES
1 2 3
1 3
1 2 3 4
1 3 4
1 1 2 3 4

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
int main() {
  int n, k;
  cin >> n >> k;
  int ar[n + 2], tr1[n + 2], tr2[n + 2], kl[k + 2][102];
  memset(kl, 0, sizeof(kl));
  vector<pair<int, int>> vec;
  for (int i = 0; i < n; i++) {
    cin >> ar[i];
    tr1[i] = ar[i];
    tr2[i] = ar[i];
  }
  sort(ar, ar + n);
  sort(tr1, tr1 + n);
  if (ar[n - 1] - ar[0] > k) {
    cout << "NO\n";
    return 0;
  }
  cout << "YES\n";
  int mn = 101;
  for (int i = 0; i < k; i++) {
    int j = 0;
    for (; j < n; j++) {
      if (ar[j] > 0) {
        mn = ar[j];
        break;
      }
    }
    for (; j < n; j++) {
      if (i == 0) {
        kl[i][j] = min(mn + 1, ar[j]);
        ar[j] -= min(mn + 1, ar[j]);
      } else {
        kl[i][j] = 1;
        ar[j] -= 1;
      }
    }
  }
  for (int i = 0; i < n; i++) {
    int f = tr2[i];
    int zz;
    for (int z = 0; z < n; z++) {
      if (tr1[z] == f) {
        zz = z;
        break;
      }
    }
    for (int j = 0; j < k; j++) {
      if (kl[j][zz] == 0) {
        continue;
      }
      for (int l = 0; l < kl[j][zz]; l++) {
        cout << j + 1 << " ";
      }
    }
    cout << "\n";
  }
  return 0;
}