B. Painting Pebbles

B. Painting Pebbles

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

There are n piles of pebbles on the table, the i-th pile contains a_i 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 b_i, 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 |b_i, c - b_j, c| ≤ 1. It isn't necessary to use all k colors: if color c hasn't been used in pile i, then b_i, 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, ..., a_n (1 ≤ a_i ≤ 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 a_i space-separated integers. j-th (1 ≤ j ≤ a_i) 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.

Sample test(s)

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

 #include <iostream>
 #include <cstdio>
 #include <cstring>
 #include <algorithm>
 #include <cmath>
 #include <string>
 #include <vector>
 #include <set>
 #include <map>
 #include <queue>
 #include <stack>
 #include <sstream>
 #include <cctype>
 #include <cassert>
 #include <typeinfo>
 #include <utility>    //std::move()
 using namespace std;
 const int INF = 0x7fffffff;
 const double EXP = 1e-;
 const int MS = ;
 typedef long long LL;

 int a[MS];
 int main()
 {
     int n, k, i, j;
     cin >> n >> k;
     int mini = INF, maxi = -INF;
     for (int i = ; i < n; i++)
     {
         cin >> a[i];
         if (mini>a[i])
             mini = a[i];
         if (maxi < a[i])
             maxi= a[i];
     }
     if ((maxi / k - )*k+ maxi%k > mini)
     {
         cout << "NO" << endl;
     }
     else
     {
         cout << "YES" << endl;
         for (i = ; i < n; i++)
         {
             for (j = ; j < a[i]; j++)
             {
                 if (j)
                     cout << " ";
                 cout << j%k + ;
             }
             cout << endl;
         }
     }
     return ;
 }