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

## 用一个list 记录所有边的信息即可
g = []
dist = [float("inf") for i in range(n+1)]

for i in range(m):
    a, b, w = map(int, input().split())
    g.append([a, b, w])

def bellman_ford():
    dist[1] = 0

    ## 循环几次代表最多用几条边
    for _ in range(k):
        ## 必须backup 防止在过程中会更新dist里面的数据导致并不是最多k条边
        backup = dist[:]
        for i in range(m):
            a, b, w = g[i][0], g[i][1], g[i][2]
            dist[b] = min(dist[b], backup[a] + w)

bellman_ford()
print(dist[n] if dist[n] != float("inf") else "impossible")