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 text {
  int x;
  int y;
} st, start;
int n, m;
int a[101];
int b[101];
int c[101][101];
queue<text> q;
int temp;
int main() {
  char s[100217 * 2];
  while (scanf("%d%d", &n, &m) != -1) {
    for (int i = 0; i < n; i++) {
      scanf("%d", &a[i]);
      b[i] = a[i];
    }
    sort(b, b + n);
    int t = b[0] / m;
    int tt = b[n - 1] / m;
    if (tt - t >= 2)
      puts("NO");
    else if (tt - t == 1) {
      if ((b[n - 1] % m > b[0] % m)) {
        puts("NO");
      } else {
        puts("YES");
        for (int i = 0; i < n; i++) {
          for (int j = 0; j < a[i] % m; j++) {
            printf(j == 0 ? "%d" : " %d", j % m + 1);
          }
          for (int j = 0; j < a[i] / m; j++) {
            for (int l = 0; l < m; l++) {
              if (a[i] % m == 0)
                printf(j == 0 && l == 0 ? "%d" : " %d", l % m + 1);
              else
                printf(" %d", l % m + 1);
            }
          }
          puts("");
        }
      }
    } else if (tt == t) {
      puts("YES");
      for (int i = 0; i < n; i++) {
        for (int j = 0; j < a[i] % m; j++) {
          printf(j == 0 ? "%d" : " %d", j % m + 1);
        }
        for (int j = 0; j < a[i] / m; j++) {
          for (int l = 0; l < m; l++) {
            if (a[i] % m == 0)
              printf(l == 0 && j == 0 ? "%d" : " %d", l % m + 1);
            else
              printf(" %d", l % m + 1);
          }
        }
        puts("");
      }
    }
  }
  return 0;
}