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 41 2 3 4

output

YES11 41 2 41 2 3 4

input

5 23 2 4 1 3

output

NO

input

5 43 2 4 3 5

output

YES1 2 31 31 2 3 41 3 41 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;}