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;
template <typename T, typename U>
inline void amin(T &x, U y) {
  if (y < x) x = y;
}
template <typename T, typename U>
inline void amax(T &x, U y) {
  if (x < y) x = y;
}
int main() {
  int n, k;
  scanf("%d%d", &n, &k);
  vector<int> a(n);
  for (int(i) = 0; (i) < (int)(n); ++(i)) scanf("%d", &a[i]);
  vector<vector<int> > ans(n);
  for (int(i) = (int)(1); (i) <= (int)(k); ++(i)) {
    int t = *min_element((a).begin(), (a).end());
    for (int(j) = 0; (j) < (int)(n); ++(j)) {
      int p = min(a[j], t + 1);
      for (int(k) = 0; (k) < (int)(p); ++(k)) {
        ans[j].push_back(i);
        --a[j];
      }
    }
  }
  if (count((a).begin(), (a).end(), 0) == n) {
    puts("YES");
    for (int(i) = 0; (i) < (int)(n); ++(i)) {
      for (int(j) = 0; (j) < (int)(ans[i].size()); ++(j)) {
        if (j != 0) putchar(' ');
        printf("%d", ans[i][j]);
      }
      puts("");
    }
  } else {
    puts("NO");
  }
  return 0;
}