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, mi = 100, ma = 0, i, j, j2, mi2, ma2;
  cin >> n >> k;
  vector<int> cnt(n);
  vector<vector<int> > clr(n);
  for (i = 0; i < n; i++) {
    cin >> cnt[i];
    mi = min(mi, cnt[i]);
    ma = max(ma, cnt[i]);
  }
  for (i = 1; i <= k; i++) {
    mi2 = 100;
    ma2 = 0;
    for (j = 0; j < n; j++) {
      bool add = cnt[j] > mi;
      for (j2 = 0; j2 < (mi + add); j2++) {
        clr[j].push_back(i);
        cnt[j]--;
      }
      mi2 = min(mi2, cnt[j]);
      ma2 = max(ma2, cnt[j]);
    }
    mi = mi2;
    ma = ma2;
  }
  if (ma)
    cout << "NO";
  else {
    cout << "YES\n";
    for (i = 0; i < n; i++) {
      for (j = 0; j < clr[i].size(); j++) cout << clr[i][j] << ' ';
      cout << '\n';
    }
  }
  return 0;
}