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;
struct ass {
  int value;
  int pos;
};
bool comp(ass a, ass b) { return a.value < b.value; }
int main() {
  int n, k;
  cin >> n >> k;
  ass *a = new ass[n];
  int max = 0;
  int min = 1000000;
  for (int i = 0; i < n; i++) {
    cin >> a[i].value;
    if (max < a[i].value) max = a[i].value;
    if (min > a[i].value) min = a[i].value;
    a[i].pos = i;
  }
  sort(a, a + n, comp);
  int **b = new int *[n];
  for (int i = 0; i < n; i++) {
    b[i] = new int[a[i].value + 1];
  }
  int z = max - min;
  int f = n;
  if (z <= k) {
    cout << "YES" << endl;
    int v = 0;
    int p = 0;
    int l = 0;
    while (n != 0) {
      l = 0;
      int m = 1;
      b[v][0] = a[v].pos;
      for (int i = 0; i <= a[v].value; i++) {
        if (i == 0) {
          for (int j = 1; j <= min; j++) {
            b[v][j] = m;
            i = j + 1;
          }
          if (v != 0) {
            b[v][i] = 1;
            i++;
            m++;
          }
        }
        if (i > a[v].value) {
          break;
        }
        if (1 == l) {
          l = 0;
          m++;
        }
        b[v][i] = m;
        l++;
      }
      n--;
      v++;
    }
    int m = 0;
    for (int x = 0; x < f; x++) {
      while (x != b[m][0]) {
        m++;
      }
      for (int i = 1; i <= a[m].value; i++) {
        cout << b[m][i] << " ";
      }
      cout << endl;
      m = 0;
    }
  } else
    cout << "NO";
}