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 suf(long long a) { return (a * (a + 1)) / 2; }
const long long OO = 1e6 + 10, MOD = 1e9 + 7;
vector<vector<int> > adj, can;
bool OK;
int main() {
  string st;
  int n, m, p, q, l, r;
  while (cin >> n >> m) {
    vector<int> v(n);
    int mx = 0, mn = INT_MAX;
    for (int i = 0; i < n; i++)
      cin >> v[i], mx = max(mx, v[i]), mn = min(mn, v[i]);
    if (mx - (mn + 1) <= m - 1) {
      cout << "YES" << endl;
      for (int i = 0; i < n; i++) {
        for (int j = 0; j < v[i] && j < mn + 1; j++) cout << 1 << " ";
        int pr = 2;
        for (int j = mn + 1; j < v[i]; j++) cout << pr++ << " ";
        cout << endl;
      }
    } else
      cout << "NO" << endl;
  }
  return 0;
}
