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 main() {
  int n, k;
  scanf("%i %i", &n, &k);
  int arr[n];
  set<pair<int, int> > piles;
  for (int i = 0; i < n; i++) {
    scanf("%i", &arr[i]);
    piles.insert(make_pair(-arr[i], i));
  }
  int ans[n][k];
  memset(ans, 0, sizeof(int) * n * k);
  int offset = 0, c = 0;
  vector<int> colours[n];
  float min1 = INT_MAX, max1 = INT_MIN;
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < arr[i]; j++) {
      colours[i].push_back((j % k) + 1);
      ans[i][j % k]++;
    }
  }
  bool flag = true;
  for (int c = 0; c < k; c++) {
    for (int i = 0; i < n; i++) {
      for (int j = i + 1; j < n; j++) {
        if (abs(ans[i][c] - ans[j][c]) > 1) {
          flag = false;
        }
      }
    }
  }
  if (!flag)
    printf("NO\n");
  else {
    printf("YES\n");
    for (int i = 0; i < n; i++) {
      for (int j = 0; j < colours[i].size(); j++) printf("%i ", colours[i][j]);
      printf("\n");
    }
  }
  return 0;
}