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() {
  vector<long long> blocks;
  vector<vector<long long>> answer;
  long long n, k;
  cin >> n >> k;
  for (long long i = 0; i < n; i++) {
    vector<long long> aux;
    answer.push_back(aux);
  }
  for (long long i = 0; i < n; i++) {
    long long aux;
    cin >> aux;
    blocks.push_back(aux);
  }
  for (long long i = 1; i <= k; i++) {
    long long minimo = LONG_MAX;
    for (long long j = 0; j < n; j++) {
      minimo = min(minimo, blocks[j]);
    }
    long long itera = 0;
    for (auto block : blocks) {
      if (block > 0) {
        if (block == minimo) {
          for (long long a = 0; a < minimo; a++) {
            answer[itera].push_back(i);
          }
        } else {
          for (long long a = 0; a < minimo + 1; a++) {
            answer[itera].push_back(i);
          }
        }
      }
      itera++;
    }
    for (long long iter = 0; iter < blocks.size(); iter++) {
      long long val = blocks[iter];
      if (val > 0) {
        if (val == minimo) {
          blocks[iter] = 0;
        } else {
          blocks[iter] = blocks[iter] - (minimo + 1);
        }
      }
    }
  }
  bool isAnswer = true;
  for (long long i = 0; i < n; i++) {
    if (blocks[i] > 0) {
      isAnswer = false;
    }
  }
  if (!isAnswer) {
    cout << "NO" << endl;
  } else {
    cout << "YES" << endl;
    for (long long i = 0; i < n; i++) {
      for (auto a : answer[i]) {
        cout << a << " ";
      }
      cout << endl;
    }
  }
  return 0;
}