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() {
  ios_base::sync_with_stdio(0);
  int n, a[101] = {0}, i, minn = 10000, b[101][101] = {0}, k, j;
  cin >> n >> k;
  for (i = 1; i <= n; i++) {
    cin >> a[i];
    minn = min(minn, a[i]);
  }
  for (i = 1; i <= n; i++) {
    if (a[i] > minn)
      b[i][1] = minn + 1;
    else
      b[i][1] = minn;
    a[i] -= b[i][1];
  }
  for (i = 1; i <= n; i++)
    for (j = 2; j - 1 <= a[i]; j++) b[i][j] = 1;
  for (i = 1; i < 101; i++)
    for (j = 1; j < 101; j++)
      if (b[i][j] > 0 && k < j) {
        cout << "NO";
        return 0;
      }
  cout << "YES"
       << "\n";
  for (i = 1; i <= n; i++) {
    for (j = 1; j < 101; j++)
      for (int l = 1; l <= b[i][j]; l++) cout << j << " ";
    cout << "\n";
  }
}