B. Painting Pebbles（Codeforces Round #289 ）

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 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.

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

代码：

#include<iostream>
#include<algorithm>
#include<stdio.h>
#include<string.h>
#include<stdlib.h>

using namespace std;

int main()
{
    int n,m;
    int a[110];
    while(scanf("%d%d",&n,&m)!=EOF)
    {
        int minn = 999999;
        int maxx = -1;
        for(int i=0;i<n;i++)
        {
            scanf("%d",&a[i]);
            if(minn > a[i])
            {
                minn = a[i];
            }
            if(maxx < a[i])
            {
                maxx = a[i];
            }
        }
        int num3 = (maxx - minn);
        if(num3>m)
        {
            printf("NO\n");
            continue;
        }
        printf("YES\n");
        for(int i=0;i<n;i++)
        {
            for(int j=0;j<a[i];j++)
            {
                if(j == 0)
                {
                    printf("%d",(j%m)+1);
                }
                else
                {
                    printf(" %d",(j%m)+1);
                }

            }
            printf("\n");
        }

    }
    return 0;
}