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(false);
  int n, k;
  cin >> n >> k;
  int a[n];
  for (int i = 0; i < n; i++) cin >> a[i];
  int mn, mx;
  int c[k][n];
  memset(c, 0, sizeof(c));
  bool flg = false;
  int lm = 0;
  for (int i = 0; i < n; i++) lm = max(lm, a[i]);
  for (int i = 0; i < lm; i++) {
    flg = false;
    for (int j = 0; j < k; j++) {
      mn = 101;
      mx = 0;
      for (int r = 0; r < n; r++) {
        if (a[r]) {
          mn = min(mn, c[j][r] + 1);
          mx = max(mx, c[j][r] + 1);
        } else {
          mn = min(mn, c[j][r]);
          mx = max(mx, c[j][r]);
        }
      }
      if ((abs(mn - mx)) <= 1) {
        flg = true;
        for (int r = 0; r < n; r++) {
          if (a[r]) {
            c[j][r] += 1;
            a[r] = a[r] - 1;
          }
        }
        break;
      }
    }
    if (!flg) break;
  }
  if (!flg)
    cout << "NO\n";
  else {
    cout << "YES\n";
    for (int j = 0; j < n; j++) {
      vector<int> tp;
      for (int i = 0; i < k; i++) {
        if (c[i][j]) {
          int cnt = c[i][j];
          while (cnt--) {
            tp.push_back(i + 1);
          }
        }
      }
      for (int i = 0; i < tp.size(); i++) {
        if (i == tp.size() - 1)
          cout << tp[i];
        else
          cout << tp[i] << " ";
      }
      cout << endl;
    }
  }
  return 0;
}