############################################################################################
import bisect,collections,copy,heapq,itertools,math,string,sys,queue,time,random
from decimal import Decimal
def I(): return input()
def IS(): return input().split()
def II(): return int(input())
def IIS(): return list(map(int,input().split()))
def LIIS(): return list(map(int,input().split()))

def make_divisors(n):
    lower_divisors , upper_divisors = [], []
    i = 1
    while i*i <= n:
        if n % i == 0:
            lower_divisors.append(i)
            if i != n // i:
                upper_divisors.append(n//i)
        i += 1
    return lower_divisors + upper_divisors[::-1]

import math

def prime_numbers(n):
    prime = [True for i in range(n+1)]
    prime[0] = False
    prime[1] = False

    sqrt_n = math.ceil(math.sqrt(n))
    for i in range(2, sqrt_n):
        if prime[i]:
            for j in range(2*i, n+1, i):
                prime[j] = False
    numbers=[]
    for i in range(2,n+1):
        if prime[i]:
            numbers.append(i)

    return numbers

def factorization(n):
    arr = []
    temp = n
    for i in range(2, int(-(-n**0.5//1))+1):
        if temp%i==0:
            cnt=0
            while temp%i==0:
                cnt+=1
                temp //= i
            arr.append([i, cnt])
 
    if temp!=1:
        arr.append([temp, 1])
 
    if arr==[]:
        arr.append([n, 1])
 
    return arr
 
INF=1<<63
MOD=998244353
MOD2=10**9+7
sys.setrecursionlimit(500005)
alpha="ABCDEFGHIJKLMNOPQRSTUVWXYZ"
def bit_count(x):
    return bin(x).count("1")
def yesno(f):
    if f:print("Yes")
    else:print("No")
####################################################
l=-1
r=10**10
n,m,x=IIS()
path=[[] for i in range(n)]
for _ in range(m):
    u,v,a,b=IIS()
    path[u-1].append((v-1,a,b))
    path[v-1].append((u-1,a,b))

def is_ok(val):
    q=[(0,0)]
    costs=[INF for i in range(n)]
    costs[0]=0
    while q:
        c,v=heapq.heappop(q)
        for i,j,k in path[v]:
            if costs[i]<c+j or val>k:continue
            heapq.heappush(q,(c+j,i))
            costs[i]=c+j
    return costs[n-1]<=x

while r-l>1:
    mid=(r+l)//2
    if is_ok(mid):
        l=mid
    else:
        r=mid
print(l)