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;
using ll = long long;
using cd = complex<double>;
ll o[(ll)(3e5 + 7)], d[(ll)(3e5 + 7)];
int main() {
  int n, k;
  cin >> n >> k;
  int ar[n];
  vector<int> vv[n];
  for (ll i = 0; i < n; i++) {
    cin >> ar[i];
    for (int j = 0; j < ar[i]; j++) {
      vv[i].push_back((j % k) + 1);
    }
  }
  sort(ar, ar + n);
  bool fl = 1;
  for (int i = 0; i < k; i++) {
    int mi = ar[0] / (k) + (i < (ar[0] % k)),
        ma = ar[n - 1] / (k) + (ar[n - 1] % k > i);
    if (ma - mi > 1) {
      fl = 0;
    }
  }
  if (!fl) {
    cout << "NO\n";
  } else {
    cout << "YES\n";
    for (int i = 0; i < n; i++) {
      for (int x : vv[i]) {
        cout << x << " ";
      }
      cout << "\n";
    }
  }
}