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.

The question ： The first line gives n,m, Next there is m Two numbers per line per line u,v, representative u Can win v, ask k The group relationship can determine the unique topological sequence , Question does not exist wei' Unique topology sequence output -1.

Ideas ： The idea of this problem is topological sorting , But be careful to record the side with the largest number , And when the number of times lost is 0 Is not unique because of the output -1, While running the topology sorting, record the side with the largest number , Pay attention to the output of special cases . The code is as follows ：

#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]++; // Record the number of times 
        }
        int ans=0;
        int tail=0,flag=1;
        for(int i=1; i<=n; i++)  // A topological sort 
        {
            if(s[i]==0)
                q[tail++]=i;
        }
        if(tail>1)flag=0;  // The value of 0 There is only one point ,flag=0 Mark the output -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) // Pay attention to j！=-1
            {
                s[e[j].v]--;
                if(s[e[j].v]==0)
                {
                    q[tail++]=e[j].v;
                    num++;
                    ans=max(ans,j); // Record number maximum side 
                }
            }
            if(num>1)
                flag=0;// The value of 0 There is only one point ,flag=0 Mark the output -1
        }
        if(flag==0||tail<n)
            printf("-1\n");  // Is not the only 
        else
            printf("%d\n",ans+1);// Topology is unique 
    }
    return 0;
}