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
#include <bits/stdc++.h>
using namespace std;
bool comp_lt(int* l, int* r) { return l[1] < r[1]; }
bool sort_piles(int* l, int* r) { return l[0] < r[0]; }
int main() {
  int n, k;
  vector<int*> arr;
  std::cin >> n >> k;
  for (int i = 0; i < n; i++) {
    int* val;
    val = new int[2];
    val[0] = i;
    cin >> val[1];
    arr.push_back(val);
  }
  vector<int*> arr_sort(arr);
  vector<int*> piles;
  sort(arr_sort.begin(), arr_sort.end(), comp_lt);
  for (int i = 0; i < n; i++) {
    int* as = new int[arr_sort[i][1] + 1];
    as[0] = arr_sort[i][0];
    piles.push_back(as);
  }
  int* used = new int[k];
  for (int i = 0; i < arr_sort[0][1]; i++) {
    piles[0][i + 1] = 1;
  }
  for (int i = 0; i < k; i++) {
    used[i] = 1;
  }
  for (int i = 1; i < n; i++) {
    for (int j = 0; j < arr_sort[i - 1][1]; j++) {
      piles[i][j + 1] = piles[i - 1][j + 1];
    }
    for (int j = arr_sort[i - 1][1]; j < arr_sort[i][1]; j++) {
      int c = 0;
      while (c < k && used[c] == 0) c++;
      if (c >= k) {
        cout << "NO";
        return 0;
      }
      piles[i][j + 1] = c + 1;
      used[c] -= 1;
    }
  }
  sort(piles.begin(), piles.end(), sort_piles);
  cout << "YES\n";
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < arr[i][1]; j++) {
      cout << piles[i][j + 1] << " ";
    }
    cout << "\n";
  }
  return 0;
}
