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;
const int MAX = 1e9;
const int MIN = -MAX;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.precision(10);
  cout << fixed;
  int _T = 1;
  for (int _t = 1; _t <= _T; _t++) {
    int n;
    int k;
    int max = MIN;
    int min = MAX;
    cin >> n >> k;
    int a[n];
    for (int i = 0; i < n; i++) {
      cin >> a[i];
      max = std::max(max, a[i]);
      min = std::min(min, a[i]);
    }
    int req = (max - (min + 1)) + 1;
    if (req > k) {
      cout << "NO\n";
      goto outer;
    }
    cout << "YES\n";
    for (int i = 0; i < n; i++) {
      int now = 1;
      for (int j = 0; j < min + 1 && a[i] > 0; j++) {
        cout << now << " ";
        a[i]--;
      }
      now = 2;
      for (int j = a[i]; j > 0; j--, now++) {
        cout << now << " ";
      }
      cout << "\n";
    }
  outer:;
  }
  return 0;
}