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 n, k, i, bolo, ae, j, l;
pair<int, int> a[1000];
vector<int> ans[1000];
int main() {
  cin >> n >> k;
  for (i = 1; i <= n; i++) {
    cin >> a[i].first;
    a[i].second = i;
  }
  sort(a + 1, a + n + 1);
  while (k--) {
    for (i = 1; i <= n; i++)
      if (a[i].first >= 0) break;
    bolo = a[i].first;
    for (j = i; j <= n; j++) {
      for (l = 1; l <= bolo + 1 && l <= a[j].first; l++)
        ans[a[j].second].push_back(k + 1);
    }
    for (j = i; j <= n; j++)
      if (a[j].first > bolo)
        a[j].first -= bolo + 1;
      else
        a[j].first -= bolo;
  }
  for (i = 1; i <= n; i++)
    if (a[i].first > 0) break;
  if (i <= n) {
    cout << "NO" << endl;
    return 0;
  }
  cout << "YES" << endl;
  for (i = 1; i <= n; i++) {
    for (j = 0; j < ans[i].size(); j++) cout << ans[i][j] << " ";
    cout << endl;
  }
}