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;
void scanint(long long int &x);
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long int n;
  scanint(n);
  long long int k;
  scanint(k);
  long long int i = 0;
  long long int a[n];
  vector<long long int> ans[n];
  long long int mini = LONG_MAX;
  long long int maxi = LONG_MIN;
  for (i = 0; i < n; i++) {
    scanint(a[i]);
    mini = min(mini, a[i]);
    maxi = max(maxi, a[i]);
  }
  if (maxi - mini > k)
    cout << "NO" << endl;
  else {
    cout << "YES" << endl;
    long long int curr = 0;
    bool b = true;
    while (b) {
      b = false;
      for (i = 0; i < n; i++) {
        if (ans[i].size() < a[i]) ans[i].push_back(curr + 1), b = true;
      }
      curr = (curr + 1) % k;
    }
    for (i = 0; i < n; i++) {
      for (int j = 0; j < ans[i].size(); j++) cout << ans[i][j] << " ";
      cout << endl;
    }
  }
  return 0;
}
void scanint(long long int &x) {
  register int c = getchar();
  x = 0;
  int neg = 0;
  for (; ((c < 48 || c > 57) && c != '-'); c = getchar())
    ;
  if (c == '-') {
    neg = 1;
    c = getchar();
  }
  for (; c > 47 && c < 58; c = getchar()) {
    x = (x << 1) + (x << 3) + c - 48;
  }
  if (neg) x = -x;
}