Robot Rapping Results Report CodeForces

问题：

While Farmer John rebuilds his farm in an unfamiliar portion of Bovinia, Bessie is out trying some alternative jobs. In her new gig as a reporter, Bessie needs to know about programming competition results as quickly as possible. When she covers the 2016 Robot Rap Battle Tournament, she notices that all of the robots operate under deterministic algorithms. In particular, robot i will beat robot j if and only if robot i has a higher skill level than robot j. And if robot i beats robot j and robot j beats robot k, then robot i will beat robot k. Since rapping is such a subtle art, two robots can never have the same skill level.

Given the results of the rap battles in the order in which they were played, determine the minimum number of first rap battles that needed to take place before Bessie could order all of the robots by skill level.

Input

The first line of the input consists of two integers, the number of robots n (2 ≤ n ≤ 100 000) and the number of rap battles m ().

The next m lines describe the results of the rap battles in the order they took place. Each consists of two integers ui and vi (1 ≤ ui, vi ≤ n, ui ≠ vi), indicating that robot ui beat robot vi in the i-th rap battle. No two rap battles involve the same pair of robots.

It is guaranteed that at least one ordering of the robots satisfies all m relations.

Output

Print the minimum k such that the ordering of the robots by skill level is uniquely defined by the first k rap battles. If there exists more than one ordering that satisfies all m relations, output -1.

Examples

Input

Output

Input

3 2
1 2
3 2

Output

-1

Note

In the first sample, the robots from strongest to weakest must be (4, 2, 1, 3), which Bessie can deduce after knowing the results of the first four rap battles.

In the second sample, both (1, 3, 2) and (3, 1, 2) are possible orderings of the robots from strongest to weakest after both rap battles.

题意：

给你n个机器人和m组比较，每组有a，b，表示a的能力比b强，问你最少通过前多少组可以确定一个唯一的拓扑排序，如果无法找到满足的结果输出-1.

思路：

首先拓扑排序每次只能找到一个能力最大了，还要n个都要找出来，然后就是找用到了第几组。

代码：

#define N 120000
#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
int n,m,f[N],book[N],b[N];
struct node
{int a,b,net;
}e[N];
int main()
{while(~scanf("%d%d",&n,&m)){memset(f,-1,sizeof(f));memset(book,0,sizeof(book));memset(b,0,sizeof(b));int l=0,flag=1,num=0,ma=-0x3f3f3f3f;for(int i=1; i<=m; i++){scanf("%d%d",&e[i].a,&e[i].b);e[i].net=f[e[i].a];//用邻接表去储存，把a打败b看作有a点到b点有条有向边f[e[i].a]=i;book[e[i].b]++;}for(int i=1; i<=n; i++)if(book[i]==0)b[l++]=i;//找到能力大的if(l!=1)flag=0;//如果能力最大的不唯一，则拓扑排序不唯一for(int i=0; i<l; i++){num=0;for(int j=f[b[i]]; j!=-1; j=e[j].net){book[e[j].b]--;//b[i]已经被排好序了，比他小的入度减少；if(!book[e[j].b])//e[i].b是当前能力最大的{b[l++]=e[j].b;//入队列，已经入队列的点表示被排好序的num++;//当前能力最大的ma=j>ma?j:ma;//记录最多用到了第几组就能有唯一的拓扑排序}}if(num>1)flag=0;//拓扑排序不唯一}if(flag==0||l<n)printf("-1\n");elseprintf("%d\n",ma);}return 0;
}

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