Codeforces Round #289 (Div. 2, ACM ICPC Rules)——B贪心——Painting Pebbles

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 Input

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

/*

   把个数从少到多进行排序，用来判断前后是否可能，只要统计循环的次数以及循环到哪里了就可以轻松判断，如果循环次数差的大于1并且差的位数也大于1那么肯定是不行的，for example:

   m = 2 a[1] = 2, a[2] = 5

   1 2   b[1] = 1  c[1] = 0;

   1 2 1 2 1 b[1] = 2 c[2] = 1

   如果循环次数差的大于2也肯定不行这个显然

   所以先判是否可以如果可以直接就根据循环和位输出当前值

*/

#include<cstdio>

#include<cstring>

#include<algorithm>

using namespace std;



struct edge{

    int a, b, c;

}G[110];



bool cmp(edge i, edge j){

    return i.a < j.a;

}

int main()

{

    int n, m;

    int a[110];

    int b[110];

    int c[110];

    while(~scanf("%d%d", &n, &m)){

        for(int i = 1; i <= n ; i++)

            scanf("%d", &a[i]);

        for(int i = 1; i <= n; i++){

            int k = a[i] / m;

            int t = a[i] % m ;

            b[i] = k;

            c[i] = t;

        }

        for(int i = 1; i <= n ;i++){

            G[i].a = a[i];

            G[i].b = b[i];

            G[i].c = c[i];

        }

        sort(G + 1, G + n + 1, cmp);

       /* for(int i = 1; i <= n ;i++){

            printf("%d %d %d", G[i].a, G[i].b, G[i].c);

           puts("");

        }

        */

        int flag = 0;

        for(int i = 1; i <= n ; i++){

            for(int j = i + 1; j <= n ; j++){

                if(G[j].b > G[i].b && G[j].c > G[i].c || G[j].b - G[i].b >= 2 ){

                   flag = 1 ;

                   break;

                }

            }

            if(flag == 1)

                break;

        }

        if(flag == 1) {printf("NO\n"); }

        else {

            printf("YES\n");

            for(int i = 1; i <= n ; i++){

                for(int j = 1; j <= b[i]; j++){

                    for(int k = 1; k <= m; k++)

                        printf("%d ",k);

                }

                for(int j = 1; j <= c[i]; j++)

                printf("%d ", j);

            puts("");

            }

        }

    }

    return 0;

}