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;
long long gcd(long long a, long long b) { return (!b) ? a : gcd(b, a % b); }
long long lcm(long long a, long long b) { return (a * b) / gcd(a, b); }
void solve() {
  long long n, k;
  cin >> n >> k;
  long long A[n];
  long long mx = INT_MIN, mn = INT_MAX;
  for (long long i = 0; i < n; ++i) {
    cin >> A[i];
    mx = max(mx, A[i]);
    mn = min(mn, A[i]);
  }
  if (mx - mn > k) {
    cout << "NO" << '\n';
    return;
  }
  cout << "YES" << '\n';
  for (long long i = 0; i < n; ++i) {
    long long ans = 0;
    for (long long j = 1; j <= A[i]; ++j) {
      cout << (ans++ % k) + 1 << ' ';
    }
    cout << '\n';
  }
}
int main() {
  std::ios_base::sync_with_stdio(false), cin.tie(nullptr), cout.tie(nullptr);
  { solve(); }
  return 0;
}