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 <class T>
T min(T a, T b, T c) {
  return min(a, min(b, c));
}
template <class T>
T min(T a, T b, T c, T d) {
  return min(a, min(b, min(c, d)));
}
template <class T>
T max(T a, T b, T c) {
  return max(a, max(b, c));
}
template <class T>
T max(T a, T b, T c, T d) {
  return max(a, max(b, max(c, d)));
}
bool cmp(const int& a, const int& b) { return a > b; }
int a[1007];
int b[1007][1007];
int main() {
  int n, k;
  scanf("%d%d", &n, &k);
  for (int i = (1); i <= (n); ++i) scanf("%d", &a[i]);
  int res = 1000000007LL;
  for (int i = (1); i <= (n); ++i) res = min(res, a[i]);
  for (int i = (1); i <= (n); ++i) {
    if (a[i] > res + k) {
      puts("NO");
      return 0;
    }
  }
  for (int i = (1); i <= (n); ++i) {
    for (int j = (1); j <= (a[i]); ++j) b[i][j] = (j - 1) % k + 1;
  }
  puts("YES");
  for (int i = (1); i <= (n); ++i) {
    for (int j = (1); j <= (a[i]); ++j) printf("%d ", b[i][j]);
    puts("");
  }
  return 0;
}