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 dr[] = {+1, -1, +0, +0};
int dc[] = {+0, +0, +1, -1};
long long gcd(long long n, long long m) { return m ? gcd(m, n % m) : n; }
long long lcm(long long n, long long m) { return n / gcd(n, m) * m; }
struct debugger {
  static void call(string::iterator it, string::iterator ed) {}
  template <typename T, typename... aT>
  static void call(string::iterator it, string::iterator ed, T a, aT... rest) {
    string b;
    for (; *it != ','; ++it)
      if (*it != ' ') b += *it;
    cout << b << "=" << a << " ";
    call(++it, ed, rest...);
  }
};
void solve() {
  int n, k;
  scanf("%d %d", &n, &k);
  ;
  int arr[n + 10];
  int mini = 1000, maxi = -1;
  for (int i = 0; i < n; i++) {
    scanf("%d", &arr[i]);
    ;
    mini = min(mini, arr[i]);
    maxi = max(maxi, arr[i]);
  }
  if (maxi - mini - k <= 0) {
    cout << "YES\n";
    for (int i = 0; i < n; i++) {
      int kk = 1;
      for (int j = 0; j < arr[i]; j++) {
        if (j <= mini) {
          cout << 1 << " ";
        } else {
          kk++;
          cout << kk << " ";
        }
      }
      cout << "\n";
    }
  } else {
    cout << "NO\n";
  }
}
int main() {
  solve();
  return 0;
}