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>
const int INF = (int)1e9;
const long long INF64 = (long long)1e18;
const long double eps = 1e-9;
const long double pi = 3.14159265358979323846;
using namespace std;
int st[150][150];
int main() {
  memset(st, 0, sizeof(st));
  int n, k;
  cin >> n >> k;
  vector<int> pile, fr;
  int tmp;
  for (int i = (0), _end = (n); i < _end; ++i) {
    cin >> tmp;
    pile.push_back(tmp);
    fr.push_back(tmp);
  }
  int ok = 1;
  vector<int> color;
  for (int i = (0), _end = (k); i < _end; ++i) {
    color.push_back(i + 1);
  }
  int cur = 0;
  int idx = 0;
  while (ok) {
    if (cur >= color.size()) cur = 0;
    int mark = 0;
    int fi = 1;
    for (int i = (0), _end = (n); i < _end; ++i) {
      if (pile[i] >= 0) {
        if (pile[i] != 0) {
          pile[i]--;
          if (pile[i] != 0) fi = 0;
          st[idx][i] = color[cur];
        } else {
          mark = 1;
        }
      }
    }
    if (fi) break;
    if (mark) {
      color.erase(color.begin() + cur);
    } else {
      cur = (cur + 1) % k;
    }
    if (color.size() == 0) {
      ok = 0;
      break;
    }
    idx++;
  }
  if (!ok)
    cout << "NO" << endl;
  else {
    cout << "YES" << endl;
    for (int i = (0), _end = (n); i < _end; ++i) {
      vector<int> re;
      for (int j = (0), _end = (fr[i]); j < _end; ++j) {
        re.push_back(st[j][i]);
      }
      sort(re.begin(), re.end());
      for (int j = (0), _end = (fr[i]); j < _end; ++j) {
        cout << re[j] << " ";
      }
      cout << endl;
    }
  }
  return 0;
}