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;
const long long int mod = 1000 * 1000 * 1000 + 7;
const double PI = 3.14159265358979323846264;
const pair<long long int, long long int> steps[] = {
    {1, 0}, {0, 1}, {-1, 0}, {0, -1}, {1, -1}, {1, 1}, {-1, -1}, {-1, 1}};
template <typename... T>
void read(T&... args) {
  ((cin >> args), ...);
}
template <typename... T>
void write(T&&... args) {
  ((cout << args << " "), ...);
}
const long long int maxN = 1e7;
void solve() {
  long long int n, k;
  read(n, k);
  vector<long long int> v(n);
  for (long long int i = 0; i < n; i++) cin >> v[i];
  ;
  for (long long int i = 0; i < n; i++) {
    for (long long int j = i + 1; j < n; j++) {
      if (abs(v[i] - v[j]) > k) {
        cout << "NO" << '\n';
        return;
      }
    }
  }
  cout << "YES" << '\n';
  for (long long int i = 0; i < n; i++) {
    long long int rep = v[i] / k;
    long long int rem = v[i] % k;
    while (rep--)
      for (long long int i = 1; i <= k; i++) {
        cout << i << " ";
      }
    for (long long int i = 1; i <= rem; i++) {
      cout << i << " ";
    }
    cout << '\n';
  }
}
void setUpLocal() {}
int32_t main() {
  cin.tie(nullptr)->sync_with_stdio(false);
  setUpLocal();
  long long int t = 1;
  while (t--) solve();
  return 0;
}