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;
const int N = 110;
int n, k, vis[N];
struct node {
  int id, ans[N], siz;
} s[N];
bool cmp1(node a, node b) { return a.siz < b.siz; }
bool cmp2(node a, node b) { return a.id < b.id; }
int main() {
  while (scanf("%d%d", &n, &k) > 0) {
    memset(s, 0, sizeof(s));
    int maxn = -1;
    for (int i = 0; i < n; i++) {
      cin >> s[i].siz;
      s[i].id = i;
      maxn = max(maxn, s[i].siz);
    }
    sort(s, s + n, cmp1);
    int flag = 0, kid = 2;
    int cnt = s[0].siz + 1;
    for (int i = 0; i < n; i++)
      for (int j = 0; j < min(cnt, s[i].siz); j++) s[i].ans[j] = 1;
    for (int i = cnt; i < maxn; i++) {
      if (kid > k) flag = 1;
      for (int j = 0; j < n; j++)
        if (s[j].siz > cnt) s[j].ans[i] = kid;
      kid++;
    }
    if (flag)
      puts("NO");
    else {
      puts("YES");
      sort(s, s + n, cmp2);
      for (int i = 0; i < n; i++) {
        for (int j = 0; j < s[i].siz; j++) cout << s[i].ans[j] << ' ';
        cout << endl;
      }
    }
  }
}