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;
int n, k;
int a[105];
int mi, ma;
int ans[105][105];
int flag;
void ini() {
  int i;
  ma = 0;
  mi = 1000;
  flag = 1;
  for (i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    ma = max(ma, a[i]);
    mi = min(mi, a[i]);
  }
}
void solve() {
  int i, j, o;
  int te = ma - mi;
  if (te > k) {
    flag = 0;
    return;
  } else {
    for (i = 1; i <= n; i++) {
      for (j = 1; j <= min(mi, a[i]); j++) {
        ans[i][j] = 1;
      }
      o = 1;
      for (; j <= a[i]; j++) {
        ans[i][j] = o;
        o++;
      }
    }
  }
}
void out() {
  int i, j;
  if (flag == 0) {
    printf("NO\n");
  } else {
    printf("YES\n");
    for (i = 1; i <= n; i++) {
      printf("%d", ans[i][1]);
      for (j = 2; j <= a[i]; j++) {
        printf(" %d", ans[i][j]);
      }
      printf("\n");
    }
  }
}
int main() {
  while (scanf("%d%d", &n, &k) != EOF) {
    ini();
    solve();
    out();
  }
  return 0;
}