import heapq
import sys
from collections import deque
input = sys.stdin.readline

INF = 10 ** 4

class Edge():
    def __init__(self, end, cost):
        self.end = end
        self.cost = cost

def dijkstra(n, G, s):
    dist = [INF] * n
    dist[s] = 0
    h = []
    heapq.heappush(h, (0, s))

    while h:
        # 使っていない頂点のうち、現時点で最も距離の近いものを選びvとする
        d, v = heapq.heappop(h)
        if dist[v] < d:
            continue
        # vから到達可能な頂点について、距離が短くなれば更新
        for e in G[v]:
            if dist[e.end] > dist[v] + e.cost:
                dist[e.end] = dist[v] + e.cost
                heapq.heappush(h, (dist[e.end], e.end))
    return dist

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

g1 = [[] for _ in range(n)]
g2 = [[] for _ in range(n)]
g3 = [[] for _ in range(n)]
g4 = [[] for _ in range(n)]

abc = []

for _ in range(m):
    a, b, c = map(int, input().split())
    abc.append((a-1, b-1, c))
    g1[a-1].append(Edge(b-1, 1))
    g2[b-1].append(Edge(a-1, 1))

dist1 = dijkstra(n, g1, 0)
dist2 = dijkstra(n, g2, n-1)

# print(dist1)
# print(dist2)

ng = set({})

for i in range(n):
    if dist1[i] == INF or dist2[i] == INF:
        ng.add(i)

H = [0] * n
cnt = 0
for a, b, c in abc:
    if a in ng or b in ng:
        continue
    cnt += 1
    g3[a].append(Edge(b, c))
    g4[b].append(Edge(a, c-p))
    H[b] += 1

stack = deque([])
topo = []

for i, hi in enumerate(H):
    if hi == 0:
        stack.append(i)

while stack:
    v = stack.pop()
    topo.append(v)
    for e in g3[v]:
        H[e.end] -= 1
        if H[e.end] == 0:
            stack.append(e.end)   

if len(topo) < cnt:
    print(1/0)
    print(-1)
    sys.exit()

f = 0
dp = ['null'] * n
dp[0] = 0

for v in topo:
    if f:
        dp[v] = 0
        for e in g4[v]:
            if e.end == 0 or dp[e.end] != "null":
                dp[v] = max(dp[v], dp[e.end] + e.cost)
    if v == 0:
        f = 1

print(dp[n-1])



### FIXED CODE ###

import heapq
import sys
from collections import deque
input = sys.stdin.readline

INF = float('inf')

class Edge():
    def __init__(self, start, end, cost):
        self.start = start
        self.end = end
        self.cost = cost

def dijkstra(n, G, s):
    dist = [INF] * n
    dist[s] = 0
    h = []
    heapq.heappush(h, (0, s))

    while h:
        # 使っていない頂点のうち、現時点で最も距離の近いものを選びvとする
        d, v = heapq.heappop(h)
        if dist[v] < d:
            continue
        # vから到達可能な頂点について、距離が短くなれば更新
        for e in G[v]:
            if dist[e.end] > dist[v] + e.cost:
                dist[e.end] = dist[v] + e.cost
                heapq.heappush(h, (dist[e.end], e.end))
    return dist

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

g1 = [[] for _ in range(n)]
g2 = [[] for _ in range(n)]
es = []

abc = []

for _ in range(m):
    a, b, c = map(int, input().split())
    abc.append((a-1, b-1, c))
    g1[a-1].append(Edge(a-1, b-1, 1))
    g2[b-1].append(Edge(b-1, a-1, 1))

dist1 = dijkstra(n, g1, 0)
dist2 = dijkstra(n, g2, n-1)

# print(dist1)
# print(dist2)

ng = set({})

for i in range(n):
    if dist1[i] == INF or dist2[i] == INF:
        ng.add(i)

for a, b, c in abc:
    if a in ng or b in ng:
        continue
    es.append(Edge(a, b, p-c))

def bellman_ford(n, es, s):
    dist = [INF] * n
    dist[s] = 0
    # 更新が止まるまで繰り返す O(V*E)
    # i度目の更新までにsを始点として0~i個の辺からなる経路が探索され、その時点の最短経路が計算される
    # 最短経路は最大でもV-1個の頂点しか通らないため、最悪でもv-1回で収束する
    while True:
        update = False
        # 辺をすべて見る
        for e in es:
            # aに到達可能かつa経由の方が近ければ更新する
            if dist[e.start] != INF and dist[e.end] > dist[e.start] + e.cost:
                dist[e.end] = dist[e.start] + e.cost
                update = True
        # 一度も更新されなければ終了
        if not update:
            break
    return dist

def find_negative_loop(n, es):
    dist = [0] * n
    for i in range(n):
        for e in es:
            if dist[e.end] > dist[e.start] + e.cost:
                dist[e.end] = dist[e.start] + e.cost
                if i == n - 1:
                    return True
    return False

if find_negative_loop(n, es):
    print(-1)
    sys.exit()

dist = bellman_ford(n, es, 0)
print(max(0, -dist[n-1]))