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 ($1 \leq m \leq \operatorname{min}(100000, \frac{n(n - 1)}{2})$).

The next m lines describe the results of the rap battles in the order they took place. Each consists of two integers u_{i} and v_{i} (1 ≤ u_{i}, v_{i} ≤ n, u_{i} ≠ v_{i}), indicating that robot u_{i} beat robot v_{i} 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
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.

Difficulty: interview

Starter code:

Solutions:
from collections import defaultdict


class RobotRapping():
    def __init__(self, n, m, battles):
        self.n, self.m = n, m
        self.battles = battles

    def generate_graph(self, k):
        edge_map = defaultdict(list)
        rev_map = defaultdict(list)
        for i in range(k):
            edge_map[self.battles[i][0]-1].append((self.battles[i][1]-1, i))
            rev_map[self.battles[i][1]-1].append((self.battles[i][0]-1, i))
        return edge_map, rev_map

    def check_order(self, num_battles):
        edge_map, rev_map = self.generate_graph(num_battles)
        outgoing_cnt = defaultdict(int)
        for k in edge_map:
            outgoing_cnt[k] = len(edge_map[k])
        s = []
        cntr = 0
        for i in range(self.n):
            if outgoing_cnt[i] == 0:
                s.append(i)
        while len(s) > cntr:
            if len(s) > cntr+1 :
                return False
            else:
                node = s[cntr]
                for v in rev_map[node]:
                    outgoing_cnt[v] -= 1
                    if outgoing_cnt[v] == 0:
                        s.append(v)
                cntr += 1
        return True

    def min_battles(self):
        if not self.check_order(self.m):
            print(-1)
        else:
            mn, mx = 0, self.m
            while mn < mx-1:
                md = int((mn+mx)/2)
                if self.check_order(md):
                    mx = md
                else:
                    mn = md
            print(mx)


    def min_battles2(self):
        edge_map, rev_map = self.generate_graph(self.m)
        outgoing_cnt = defaultdict(int)
        for k in edge_map:
            outgoing_cnt[k] = len(edge_map[k])
        s = []
        cntr = 0
        order = []
        for i in range(self.n):
            if outgoing_cnt[i] == 0:
                s.append(i)
        while len(s) > cntr:
            if len(s) > cntr+1 :
                print(-1)
                return
            else:
                node = s[cntr]
                order.append(node)
                for v,_ in rev_map[node]:
                    outgoing_cnt[v] -= 1
                    if outgoing_cnt[v] == 0:
                        s.append(v)
                cntr += 1
        mn_pos = -1
        for i in range(1,self.n):
            for v,ind in edge_map[order[i]]:
                if v == order[i-1]:
                    mn_pos = max(mn_pos, ind)
                    break
        print(mn_pos+1)

n,m = list(map(int,input().strip(' ').split(' ')))
battles = []
for i in range(m):
    x,y = list(map(int,input().strip(' ').split(' ')))
    battles.append((x,y))
rr = RobotRapping(n,m,battles)
rr.min_battles2()


from heapq import heappop,heappush

n,m = list(map(int,input().split()))

C = [[] for _ in range(n)]
indeg = [0]*n

def toposort():
    S = [i for i in range(n) if indeg[i] == 0]
    nparent = indeg[:]
    topo = []
    while S:
        cur = S.pop()
        topo.append(cur)
        for neigh,_ in C[cur]:
            nparent[neigh] -= 1
            if nparent[neigh] == 0:
                S.append(neigh)
    return topo

def solve():
    topo = toposort()

    D = [(0,0)]*n
    for cur in topo:
        for neigh,t in C[cur]:
            cd,ct = D[cur]
            nd,_ = D[neigh]
            if nd <= cd + 1:
                D[neigh] = cd + 1, max(ct,t)

    d,t = max(D)
    return t+1 if d == n-1 else -1
        

for _ in range(m):
    a,b = list(map(int,input().split()))
    C[a-1].append((b-1, _))
    indeg[b-1] += 1

print(solve())


n, m = [int(i) for i in input().split()]
n += 1 # one-indexed
A = [[int(i) for i in input().split()] for j in range(m)]
m += 1

def check(upper):
    p = [[] for i in range(n)]
    d = [0] * n #record num of parents
    for u, v in A[:upper]:
        p[u].append(v) # form arc from u to v
        d[v] += 1
    if d.count(0) > 2:
        return False
    x = d.index(0, 1) # find the real ancestor, should only be one
    while x:
        q, x = p[x], 0
        for y in q:
            d[y] -= 1
            if d[y] == 0:
                if x: 
                    return False
                x = y
    return True

left, right = 1, m
while left < right:
    mid = (left + right)//2
    if check(mid): 
        right = mid
    else:
        left = mid + 1 
if check(left):
    print(left)
else:
    print(-1)


n, m = [int(i) for i in input().split()]
n += 1 # one-indexed
A = [[int(i) for i in input().split()] for j in range(m)]
m += 1

def check(upper):
    p = [[] for i in range(n)]
    d = [0] * n #record num of parents
    for u, v in A[:upper]:
        p[u].append(v) # form arc from u to v
        d[v] += 1
    if d.count(0) > 2:
        return False
    x = d.index(0, 1) # find the real ancestor, should only be one
    while x:
        q, x = p[x], 0
        for y in q:
            d[y] -= 1
            if d[y] == 0:
                if x: 
                    return False
                x = y
    return True

left, right = 1, m
while left < right:
    mid = (left + right)//2
    if check(mid): 
        right = mid
    else:
        left = mid + 1 
if check(left):
    print(left)
else:
    print(-1)

from sys import stdin, stdout
import sys
import bisect
import heapq
input = sys.stdin.readline


def solve(n, m, edges):
    
    lo = 0
    hi = m

    curr_k = -1

    while lo <= hi:
        mid = lo + (hi - lo) // 2

        can_do = True

        # condition

        adj_list = {x: [] for x in range(0, n)}
        in_degree = [0] * n
        for ed in range(min(mid, len(edges))):
            edge = edges[ed]
            adj_list[edge[0]].append(edge[1])
            in_degree[edge[1]] += 1
        candidates = []
        for i in range(len(in_degree)):
            if in_degree[i] == 0:
                candidates.append(i)
        res = []
        while candidates:
            ele = candidates.pop(0)
            if len(candidates) > 0:
                can_do = False
                break
            res.append(ele)
            for i in adj_list[ele]:
                in_degree[i] -= 1
                if in_degree[i] == 0:
                    candidates.append(i)
        
        if len(res) < n:
            can_do = False

        if can_do:
            curr_k = mid
            hi = mid - 1
        else:
            lo = mid + 1

    return curr_k

def main():
    n, m = list(map(int, input().split()))
    
    edges = []
    for i in range(m):
        a, b = list(map(int, input().split()))
        edges.append([a - 1, b - 1])
    stdout.write(str(solve(n, m, edges)))
    stdout.write("\n")


def __starting_point():
    main()

__starting_point()