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 cnt[1000000 + 1];
const int mod = 1e9 + 7;
void solve() {
  int n, k;
  cin >> n >> k;
  int a[n], b[n];
  int r;
  vector<int> v;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    b[i] = a[i];
  }
  sort(a, a + n);
  r = a[0] / k;
  int d = 0;
  int mn[101] = {0};
  int mx[101] = {0};
  if (a[0] % k != 0) d = 1;
  for (int i = 1; i <= k; i++) {
    mn[i] += r;
  }
  if (d == 1) {
    int m = a[0] % k;
    for (int i = 1; m--; i++) {
      mn[i] += 1;
    }
  }
  r = a[n - 1] / k;
  d = 0;
  if (a[n - 1] % k != 0) d = 1;
  for (int i = 1; i <= k; i++) {
    mx[i] += r;
  }
  if (d == 1) {
    int m = a[n - 1] % k;
    for (int i = 1; m--; i++) {
      mx[i] += 1;
    }
  }
  for (int i = 1; i <= 100; i++) {
    if (abs(mn[i] - mx[i]) > 1) {
      cout << "NO" << endl;
      return;
    }
  }
  cout << "YES" << endl;
  for (int i = 0; i < n; i++) {
    int d = 1;
    int x = b[i];
    while (x--) {
      cout << d << " ";
      d++;
      if (d > k) d = 1;
    }
    cout << endl;
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  int t;
  t = 1;
  while (t--) solve();
  return 0;
}