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;
}