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
#include <bits/stdc++.h>
using namespace std;
long long int gcd(long long int a, long long int b) {
  if (a == 0) return b;
  return gcd(b % a, a);
}
int main() {
  long long int t, b, m, n, i, j, temp, mm, mx, hell, l, r, k, ans, c, p, q,
      val, count, total, sum;
  long long int a[100001];
  vector<long long int> v[103];
  string s;
  cin >> n >> k;
  for (i = 0; i < n; i++) cin >> a[i];
  for (i = 0; i < n; i++) {
    temp = 1;
    for (j = 0; j < a[i]; j++) {
      v[i].push_back(temp);
      temp++;
      if (temp > k) temp = temp - k;
    }
  }
  mm = 1000000000;
  mx = 0;
  int f = 0;
  for (p = 1; p <= k; p++) {
    mm = 1000000000;
    mx = 0;
    for (i = 0; i < n; i++) {
      if (a[i] % k >= p)
        val = a[i] / k + 1;
      else
        val = a[i] / k;
      mm = min(mm, val);
      mx = max(mx, val);
    }
    if ((mx - mm) > 1) f = 1;
  }
  if (f == 1)
    cout << "NO" << endl;
  else {
    cout << "YES" << endl;
    for (i = 0; i < n; i++) {
      for (j = 0; j < v[i].size(); j++) cout << v[i][j] << " ";
      cout << endl;
    }
  }
  return 0;
}
