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;
vector<pair<int, int> > A;
map<pair<int, int>, vector<int> > M;
int main() {
  int n, c;
  scanf("%d", &n);
  scanf("%d", &c);
  for (int i = (0); i < (n); ++i) {
    int a;
    scanf("%d", &a);
    A.push_back(pair<int, int>(a, i));
  }
  sort(A.begin(), A.end());
  int flag = 0;
  for (int i = (0); i < (c); ++i) {
    if (flag == 1) {
      for (int j = (1); j < (n); ++j) {
        if (M[A[j]].size() != A[j].first) {
          M[A[j]].push_back(i);
        }
      }
    } else {
      for (int j = (0); j < (n); ++j) {
        for (int k = (0); k < (A[0].first); ++k) {
          M[A[j]].push_back(i);
        }
        if (A[j].first > A[0].first) {
          M[A[j]].push_back(i);
        }
      }
      flag = 1;
    }
  }
  if (M[A[n - 1]].size() != A[n - 1].first) {
    cout << "NO" << endl;
    return 0;
  } else {
    cout << "YES" << endl;
    vector<int> ans[n];
    for (int i = (0); i < (n); ++i) {
      ans[A[i].second] = M[A[i]];
    }
    for (int i = (0); i < (n); ++i) {
      for (int j = (0); j < (ans[i].size()); ++j) {
        cout << ans[i][j] + 1 << " ";
      }
      cout << endl;
    }
  }
  return 0;
}