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>
inline void checkmin(T &a, T b) {
  if (a > b) a = b;
}
template <class T>
inline void checkmax(T &a, T b) {
  if (a < b) a = b;
}
const double eps = 1e-7;
const int N = 1;
const int M = 200000;
const long long P = 10000000097ll;
const int INF = 0x3f3f3f3f;
int a[110][110];
int main() {
  int i, j, t, k, n, m, v, u, numCase = 0;
  int MAX = -INF, MIN = INF;
  cin >> n >> k;
  for (i = 1; i <= n; ++i) {
    cin >> a[i][0];
    checkmin(MIN, a[i][0]);
    checkmax(MAX, a[i][0]);
  }
  if (MAX - MIN > k) {
    cout << "NO" << endl;
    return 0;
  }
  cout << "YES" << endl;
  for (i = 1; i <= n; ++i) {
    int num = a[i][0];
    for (j = 1; j <= MIN; ++j) {
      cout << '1' << ' ';
    }
    int delta = num - MIN;
    j = 1;
    while (delta) {
      cout << j++ << ' ';
      delta--;
    }
    cout << endl;
  }
}