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 maxn = 124;
int m, n, brd[maxn][maxn], minn, maxx;
bool solve() {
  for (int i = 0; i < m; i++) {
    int len = min(minn, brd[i][0]), num = 2, j;
    for (j = 1; j <= minn; j++) brd[i][j] = 1;
    while (j <= brd[i][0]) {
      brd[i][j] = num++;
      j++;
    }
  }
  return true;
}
int main() {
  while (~scanf("%d%d", &m, &n)) {
    minn = 0x7f7f7f7f;
    maxx = 0;
    memset(brd, 0, sizeof(brd));
    for (int i = 0; i < m; i++) {
      scanf("%d", &brd[i][0]);
      minn = min(minn, brd[i][0]);
      maxx = max(maxx, brd[i][0]);
    }
    minn++;
    if (maxx - minn + 1 > n) {
      printf("NO\n");
      continue;
    }
    solve();
    printf("YES\n");
    for (int i = 0; i < m; i++) {
      for (int j = 1; j <= brd[i][0]; j++) {
        if (j > 1) printf(" ");
        printf("%d", brd[i][j]);
      }
      printf("\n");
    }
  }
  return 0;
}