// 間違っていたので修正
// O(N^2M) 解法

#include<iostream>
#include<algorithm>
#include<vector>
using namespace std;
using ll = long long;


bool dp[1000000][100];
int frm[10000];
int to[10000];

ll N, M, T, ans;

int main(){
    cin >> N >> M >> T;
    for(int i=0;i<M;i++){
        int from[i], b;
        cin >> frm[i] >> to[i];
    }
    dp[0][0] = true;
    int T_max = N * N + 10 * N;
    for(int t=1;t<=T_max;t++){
        for(int e=0;e<M;e++){
            dp[t][to[e]] |= dp[t-1][frm[e]];
        }
    }
    for(int v=0;v<N;v++){
        // last two
        ll t1 = -1;
        ll t2 = -1;
        for (int t=0;t<T_max;t++){
            if(dp[t][v]){
                t1 = t2;
                t2 = t;
            }
        }
        if(T < T_max){
            ans += (dp[T][v] ? 1 : 0);
        } else{
            if(t1 > N){
                ll p = t2 - t1;
                ans += ((T-t2)%p==0 ? 1 : 0);
            }
        }
    }
    if(ans==0){
        ans-=1;
    }
    cout << ans << endl;
    return 0;
}