Codeforces 417C Football【思维】

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本文介绍了一种足球赛程安排算法，确保每个队伍在比赛中恰好获胜K次，并满足每两队仅比赛一次的条件。通过构造合理的比赛序列，解决了由Pavel发起的足球友谊赛结果推演问题。

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One day, at the "Russian Code Cup" event it was decided to play football as an out of competition event. All participants was divided into n teams and played several matches, two teams could not play against each other more than once.

The appointed Judge was the most experienced member — Pavel. But since he was the wisest of all, he soon got bored of the game and fell asleep. Waking up, he discovered that the tournament is over and the teams want to know the results of all the matches.

Pavel didn't want anyone to discover about him sleeping and not keeping an eye on the results, so he decided to recover the results of all games. To do this, he asked all the teams and learned that the real winner was friendship, that is, each team beat the other teams exactly k times. Help Pavel come up with chronology of the tournir that meets all the conditions, or otherwise report that there is no such table.

Input

The first line contains two integers — n and k (1 ≤ n, k ≤ 1000).

Output

In the first line print an integer m — number of the played games. The following m lines should contain the information about all the matches, one match per line. The i-th line should contain two integers a_i and b_i (1 ≤ a_i, b_i ≤ n; a_i ≠ b_i). The numbers a_i and b_i mean, that in the i-th match the team with number a_i won against the team with number b_i. You can assume, that the teams are numbered from 1 to n.

If a tournir that meets the conditions of the problem does not exist, then print -1.

Examples

Input

3 1

Output

题目大意：

一共有N个球队，已知每个球队获胜的场次都是K次，问怎样分配比赛使得每个队都能获胜K次，且每两个队之间至多进行一场比赛。

思路：

将比赛可能进行的场次看成无向边，那么对应一个有N个球队的完全无向图的边数为n*(n-1)/2条，也就是最多有这些比赛场次。

对应我们知道，每个队伍能够进行的比赛数都是n-1场，那么平均分配一下即可，若k*n<=(n-1)*n/2.那么肯定存在解，我们对应将每个队伍的获胜场次都控制在K次即可。

否则无解。

Ac代码：

#include<stdio.h>
#include<string.h>
using namespace std;
int degree[4000];
int main()
{
    int n,k;
    while(~scanf("%d%d",&n,&k))
    {
        memset(degree,0,sizeof(degree));
        if((n*(n-1)/2)>=n*k)
        {
            printf("%d\n",n*k);
            for(int i=1;i<=n;i++)
            {
                for(int j=i;j<=n;j++)
                {
                    if(i==j)continue;
                    if(degree[i]<k)
                    {
                        printf("%d %d\n",i,j);degree[i]++;
                    }
                    else if(degree[j]<k)
                    {
                        printf("%d %d\n",j,i);degree[j]++;
                    }
                    else continue;
                }
            }
        }
        else printf("-1\n");
    }
}