Robot Rapping Results Report

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 viin 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

4 5
2 1
1 3
2 3
4 2
4 3

Output

4

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：第一行给定n，m，接下来有m行每行每行两个数u，v，代表u能赢v,问k组关系就能确定唯一的拓扑序列，问不存在wei'唯一的拓扑序列输出-1.

思路：这个题的思想是拓扑排序，但注意要记录编号最大的边，并且当输的次数为0的情况不唯一时因该输出-1，跑一边拓扑排序同时记录一下编号最大的边，注意特殊情况输出即可。代码如下：

#include <stdio.h>
#include <string.h>
#include <queue>
#define maxn 100005
using namespace std;
using namespace std;
struct node
{
    int u,v,next;
} e[100010];
int head[100010],cnt;
int n,m;
int q[100010];
int s[100010];
int main()
{
    while(~scanf("%d %d",&n,&m))
    {
        memset(head,-1,sizeof(head));
        cnt=0;
        memset(s,0,sizeof(s));
        for(int i=0; i<m; i++)
        {
            int u,v;
            scanf("%d%d",&u,&v);
            e[cnt].u=u;
            e[cnt].v=v;
            e[cnt].next=head[u];
            head[u]=cnt++;
            s[v]++; //记录输的次数
        }
        int ans=0;
        int tail=0,flag=1;
        for(int i=1; i<=n; i++)  //拓扑排序
        {
            if(s[i]==0)
                q[tail++]=i;
        }
        if(tail>1)flag=0;  //这个值为0的点只有一个，flag=0标记该输出-1
        for(int i=0; i<tail; i++)
        {
            int t=q[i];
            int num=0;
            for(int j=head[t]; j!=-1; j=e[j].next) //注意就j！=-1
            {
                s[e[j].v]--;
                if(s[e[j].v]==0)
                {
                    q[tail++]=e[j].v;
                    num++;
                    ans=max(ans,j); //记录编号最大边
                }
            }
            if(num>1)
                flag=0;//这个值为0的点只有一个，flag=0标记该输出-1
        }
        if(flag==0||tail<n)
            printf("-1\n");  //不唯一
        else
            printf("%d\n",ans+1);//拓扑唯一
    }
    return 0;
}