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 Set(int N, int pos) { return N = N | (1 << pos); }
int reset(int N, int pos) { return N = N & ~(1 << pos); }
bool check(int N, int pos) { return (bool)(N & (1 << pos)); }
struct edge {
  int w, s;
};
bool cmp(const edge &a, const edge &b) { return a.w < b.w; }
long long gcd(long long a, long long b) { return !a ? b : gcd(b % a, a); }
long long lcm(long long a, long long b) { return a / gcd(a, b) * b; }
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int n, k, d, a[123], mn = INT_MAX, mx = -1;
  cin >> n >> k;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    mn = min(mn, a[i]);
    mx = max(mx, a[i]);
  }
  if (k >= mx) {
    cout << "YES\n";
    for (int i = 0; i < n; i++) {
      for (int j = 1; j < a[i] + 1; j++) cout << j << " ";
      cout << endl;
    }
  } else if (k + mn < mx)
    cout << "NO";
  else {
    cout << "YES\n";
    for (int i = 0; i < n; i++) {
      if (a[i] <= k) {
        for (int j = 1; j < a[i] + 1; j++) cout << j << " ";
        cout << endl;
      } else {
        int di = a[i] / k;
        int dm = a[i] % k;
        for (int l = 0; l < di; l++) {
          for (int j = 1; j < k + 1; j++) cout << j << " ";
        }
        for (int j = 1; j < dm + 1; j++) cout << j << " ";
        cout << endl;
      }
    }
  }
  return 0;
}