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;
struct node {
  int num, id;
  int a[110];
} p[110];
int vis[110];
int main() {
  int n, k, i, j, l, temp, num, flag;
  scanf("%d %d", &n, &k);
  for (i = 0; i < n; i++) {
    scanf("%d", &p[i].num);
    p[i].id = i;
    memset(p[i].a, 0, sizeof(p[i].a));
  }
  memset(vis, 0, sizeof(vis));
  temp = 1;
  while (1) {
    flag = 0;
    for (i = 0; i < n; i++) {
      if (p[i].num) {
        p[i].num--;
        p[i].a[temp]++;
      } else
        flag++;
    }
    if (flag == n) break;
    if (flag) vis[temp] = 1;
    flag = temp;
    temp++;
    if (temp == k + 1) temp = 1;
    while (vis[temp] == 1) {
      temp++;
      if (temp == k + 1) temp = 1;
      if (temp == flag) break;
    }
    if (vis[temp]) break;
  }
  for (i = 0; i < n; i++)
    if (p[i].num) break;
  if (i < n) {
    printf("NO\n");
  } else {
    printf("YES\n");
    for (i = 0; i < n; i++) {
      flag = 0;
      for (j = 1; j <= k; j++) {
        if (p[i].a[j]) {
          for (l = 0; l < p[i].a[j]; l++) {
            if (flag == 0) {
              printf("%d", j);
              flag = 1;
            } else {
              printf(" %d", j);
            }
          }
        }
      }
      printf("\n");
    }
  }
  return 0;
}
