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;
static int n, k;
static vector<int> v;
int main() {
  while (scanf("%d%d", &n, &k) != EOF) {
    v.clear();
    v.resize(n);
    int maxinum = INT_MIN, minium = INT_MAX;
    for (int i = 0; i < n; i++) {
      scanf("%d", &v[i]);
      maxinum = max(v[i], maxinum);
      minium = min(v[i], minium);
    }
    if (maxinum - minium > k) {
      printf("NO\n");
    } else {
      printf("YES\n");
      for (int i = 0; i < n; i++) {
        for (int j = 1; j <= v[i] && j <= k; j++) printf("%d ", j);
        if (k < v[i]) {
          int cnt = 0;
          while (cnt < v[i] - k) {
            for (int j = 1; j <= k; j++) {
              printf("%d ", j);
              ++cnt;
              if (cnt == v[i] - k) break;
            }
          }
        }
        printf("\n");
      }
    }
  }
  return 0;
}