B. Painting Pebbles

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 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 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

#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-8;
const int MS = 101;
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 = 0; i < n; i++)
    {
        cin >> a[i];
        if (mini>a[i])
            mini = a[i];
        if (maxi < a[i])
            maxi= a[i];
    }
    if ((maxi / k - 1)*k+ maxi%k > mini)
    {
        cout << "NO" << endl;
    }
    else
    {
        cout << "YES" << endl;
        for (i = 0; i < n; i++)
        {
            for (j = 0; j < a[i]; j++)
            {
                if (j)
                    cout << " ";
                cout << j%k + 1;
            }
            cout << endl;
        }
    }
    return 0;
}