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.

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.

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.

4 4
1 2 3 4

YES
1
1 4
1 2 4
1 2 3 4

5 2
3 2 4 1 3

NO

5 4
3 2 4 3 5

 #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 ;

 }