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
#include <bits/stdc++.h>
using namespace std;
const int N = 105;
int b[N][N], a[N], n, k;
bool kt(int* a) {
  for (int i = 1; i <= n; ++i)
    if (a[i]) return true;
  return false;
}
bool check(int i, int j) {
  for (int l = 1; l <= n; ++l)
    if (abs(b[l][j] - b[i][j] - 1) > 1) return false;
  return true;
}
int main() {
  scanf("%d%d", &n, &k);
  for (int i = 1; i <= n; ++i) scanf("%d", &a[i]);
  while (kt(a)) {
    bool f = false;
    for (int i = 1; i <= n; ++i) {
      if (a[i])
        for (int j = 1; j <= k; ++j) {
          if (!check(i, j)) continue;
          f = true;
          b[i][j]++;
          a[i]--;
          break;
        }
    }
    if (!f) {
      printf("NO");
      return 0;
    }
  }
  printf("YES\n");
  for (int i = 1; i <= n; ++i) {
    for (int j = 1; j <= k; ++j)
      for (int t = 1; t <= b[i][j]; ++t) printf("%d ", j);
    printf("\n");
  }
  return 0;
}
