#define _USE_MATH_DEFINES
#include <iostream>
#include <fstream>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <cstring>
#include <cassert>
#include <string>
#include <vector>
#include <utility>
#include <complex>
#include <set>
#include <map>
#include <unordered_map>
#include <queue>
#include <stack>
#include <deque>
#include <tuple>
#include <bitset>
#include <limits>
#include <algorithm>
#include <array>
#include <random>
#include <complex>
#include <regex>
using namespace std;
typedef long double ld;
typedef long long ll;
typedef vector<int> vint;
typedef vector<ll> vll;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
typedef pair<double, double> pdd;
typedef complex<ld> compd;
#define quickIO()	{cin.tie(0);	cout.sync_with_stdio(false);}
#define reach(i,a)	for(auto i:a)
#define rep(i,n)	for(int i=0;i<((int)n);i++)
#define REP(i,n)	for(int i=0;i<=((int)n);i++)
#define srep(i,a,n)	for(int i=a;i<((int)n);i++)
#define SREP(i,a,n)	for(int i=a;i<=((int)n);i++)
#define rrep(i,n)	for(int i=n-1;i>=0;i--)
#define RREP(i,n)	for(int i=n;i>=0;i--)
#define all(a)	(a).begin(),(a).end()
#define mp(a,b)	make_pair(a,b)
#define mt	make_tuple
#define pb	push_back
template<typename T> istream & operator >> (istream & is, vector<T> & vec) {
	for (T& x : vec)	is >> x;
	return is;
}
template<typename T> ostream& operator << (ostream & os, vector<T> & vec) {
	os << "[";
	rep(i, vec.size())	os << (i ? ", " : "") << vec[i];
	os << "]";
	return os;
}
template<typename T> istream & operator >> (istream & is, pair<T, T> & p) {
	is >> p.first >> p.second;
	return is;
}
template<typename T> ostream& operator << (ostream & os, pair<T, T> & p) {
	os << p.first << " " << p.second;
	return os;
}
int bitcnt(ll x) {
	x = ((x & 0xAAAAAAAAAAAAAAAA) >> 1) + (x & 0x5555555555555555);
	x = ((x & 0xCCCCCCCCCCCCCCCC) >> 2) + (x & 0x3333333333333333);
	x = ((x & 0xF0F0F0F0F0F0F0F0) >> 4) + (x & 0x0F0F0F0F0F0F0F0F);
	x = ((x & 0xFF00FF00FF00FF00) >> 8) + (x & 0x00FF00FF00FF00FF);
	x = ((x & 0xFFFF0000FFFF0000) >> 16) + (x & 0x0000FFFF0000FFFF);
	x = ((x & 0xFFFFFFFF00000000) >> 32) + (x & 0x00000000FFFFFFFF);
	return x;
}
int bitcnt(int x) {
	x = ((x & 0xAAAAAAAA) >> 1) + (x & 0x55555555);
	x = ((x & 0xCCCCCCCC) >> 2) + (x & 0x33333333);
	x = ((x & 0xF0F0F0F0) >> 4) + (x & 0x0F0F0F0F);
	x = ((x & 0xFF00FF00) >> 8) + (x & 0x00FF00FF);
	x = ((x & 0xFFFF0000) >> 16) + (x & 0x0000FFFF);
	return x;
}
ll sqrtll(ll x) {
	ll left = 0, right = x;
	rep(i, 100) {
		ll mid = (left + right) >> 1;
		if (mid * mid <= x)	left = mid;
		else	right = mid;
	}
	return left;
}
ll gcd(ll a, ll b) {
	return b == 0 ? a : gcd(b, a % b);
}
#define debug(x)	printf("Case #%d: ", x)
#define DEBUG 0
const ld infl = 1e100;
const ll mod = 1e9 + 7;
const ld eps = 1e-9;
const ll inf = 1e15;
const int dx[] = { 1,0,-1,0,0 };
const int dy[] = { 0,1,0,-1,0 };

vector<tuple<ll, ll, ll>> edge;
bool updated_nodes[3000];
ll dist[3000];

// 各辺のコストをP-(コイン)と置き換えると最短距離問題(答えは-1をかける)になりそう
// 無限ループあるのでベルマンフォードでN回目と2N回目で答えが変わっているか検証してね

void bellman_ford(int n, int m) {
	// n回辺の更新をします
	rep(i, n) {
		bool updated = false;
		rep(j, m) {
			ll a, b, c;
			tie(a, b, c) = edge[j];
			if (dist[a] == -inf)	continue;
			if (dist[b] < dist[a] + c) {
				dist[b] = dist[a] + c;
				updated = true;
			}
		}
		if (updated == false)	break;
	}
}

int main() {
	ll n, m, p;	cin >> n >> m >> p;
	rep(i, m) {
		int a, b, c;	cin >> a >> b >> c;	a--;	b--;
		edge.push_back(mt(a, b, c - p));
	}

	// 初期化
	rep(i, n)	dist[i] = -inf;
	dist[0] = 0;
	
	bellman_ford(n, m);
	ll res1 = dist[n - 1];
	rep(i, n) {
		rep(j, m) {
			ll a, b, c;	tie(a, b, c) = edge[j];
		if (dist[a] == -inf)	continue;
			if (dist[b] < dist[a] + c) {
				updated_nodes[b] = true;
			}
			else if (updated_nodes[a]) {
				updated_nodes[b] = true;
			}	
		}
	}


	if (updated_nodes[n - 1]) {
		cout << -1 << endl;
	}
	else {
		cout << max(0LL, res1) << endl;
	}
	return 0;
}

Measured memory_bytes: 576