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 ans[105][105], n, m, i, j, k;
struct node {
  int x, y;
} t[105];
int cmp(node i, node j) { return i.x < j.x; }
int main() {
  scanf("%d%d", &n, &m);
  for (i = 1; i <= n; i++) scanf("%d", &t[i].x);
  for (i = 1; i <= n; i++) t[i].y = i;
  sort(t + 1, t + n + 1, cmp);
  for (i = 1; i <= m; i++) {
    for (j = 1; j < n; j++)
      if (t[j + 1].x > t[j].x) break;
    for (k = 1; k <= j; k++) {
      ans[t[k].y][i] = t[1].x;
    }
    for (k = j + 1; k <= n; k++) ans[t[k].y][i] = t[1].x + 1;
    for (k = j + 1; k <= n; k++) t[k].x -= (t[1].x + 1);
    for (k = j; k >= 1; k--) t[k].x -= t[1].x;
  }
  if (t[n].x) {
    cout << "NO";
    return 0;
  }
  cout << "YES" << endl;
  for (i = 1; i <= n; i++) {
    for (j = 1; j <= m; j++)
      while (ans[i][j]) {
        cout << j << ' ';
        ans[i][j]--;
      }
    cout << endl;
  }
  return 0;
}