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 i, j, l, s, k, n;
  cin >> n >> k;
  l = 0;
  s = 100000;
  int a[n];
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    if (a[i] > l) l = a[i];
    if (a[i] < s) s = a[i];
  }
  int xx;
  if (k < s)
    xx = 2 * k + s - k;
  else
    xx = 2 * s + k - s;
  if ((l) <= xx) {
    cout << "YES" << endl;
    if (k >= s) {
      for (int i = 0; i < n; i++) {
        if (a[i] == s) {
          for (int j = 0; j < a[i]; j++) cout << j + 1 << " ";
        } else {
          for (int j = 0; j < a[i]; j++) {
            if (j < s)
              cout << j + 1 << " ";
            else if (j < 2 * s)
              cout << j - s + 1 << " ";
            else
              cout << j - 2 * s + s + 1 << " ";
          }
        }
        cout << endl;
      }
    } else {
      for (int i = 0; i < n; i++) {
        int yy = s / k;
        for (int j = 0; j < a[i]; j++) {
          if (j < yy * k)
            cout << j / yy + 1 << " ";
          else {
            if ((j - yy * k + 1) > k)
              cout << (j - yy * k + 1) - k << " ";
            else
              cout << (j - yy * k + 1) << " ";
          }
        }
        cout << endl;
      }
    }
  } else
    cout << "NO";
  return 0;
}