//存在负权值   处理负环

//如果能求出来  一般是不存在负权回路

//如果有负回路  那最小距离可能是负无穷

#include <cstring>

#include <iostream>

using namespace std;

const int N = 1e4+;

struct Edge {

    int a;

    int b;

    int w;

} edge[N];

int n, m, k;

int dist[N], backup[N];

void bellman_ford() {

    memset(dist, 0x3f, sizeof dist);

    dist[] = ;

    ////迭代k次，表示经过不超过k条边走到每个点的距离

    for (int i=; i<k; i++) {

        memcpy(backup , dist, sizeof dist);//备份，不加备份可能出现串联

        for (int j=; j<m; j++) {

            int a=edge[j].a,b=edge[j].b,w=edge[j].w;

            dist[b] = min(dist[b],backup[a]+w);//只用上一次的

        }

    }

    if (dist[n] > 0x3f3f3f3f/) cout << "impossible";

    else cout << dist[n];

}

int main() {

    cin >> n >> m >> k;

    for (int i=; i<m; i++) {

        int a, b, w;

        cin >> a >> b >> w;

        edge[i] = {a, b, w};

    }

    bellman_ford();

    return ;

}