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;
struct Pile {
  int w, id;
  int c[N], cn;
  bool operator<(const Pile& rhs) const { return w < rhs.w; }
} a[N];
bool cmpid(const Pile& a, const Pile& b) { return a.id < b.id; }
int main() {
  scanf("%d%d", &n, &k);
  for (int i = 0; i < n; i++) {
    scanf("%d", &a[i].w);
    a[i].id = i;
  }
  sort(a, a + n);
  int nc = 0;
  for (int i = 0; i < n; i++) {
    while (a[i].cn < a[i].w) {
      nc++;
      if (nc > k) {
        puts("NO");
        return 0;
      }
      int rnd = 1;
      if (i == 0) rnd = a[i].w;
      while (rnd--) {
        for (int j = i; j < n; j++) {
          a[j].c[a[j].cn++] = nc;
        }
      }
      if (i == 0) {
        for (int j = i + 1; j < n; j++) {
          if (a[j].cn < a[j].w) a[j].c[a[j].cn++] = nc;
        }
      }
    }
  }
  sort(a, a + n, cmpid);
  puts("YES");
  for (int i = 0; i < n; i++) {
    printf("%d", a[i].c[0]);
    for (int j = 1; j < a[i].cn; j++) {
      printf(" %d", a[i].c[j]);
    }
    puts("");
  }
  return 0;
}