#include <iostream>
#include <cstring>
#include <queue>

using namespace std;

typedef pair<int, int> PII;

const int N = 1000010;

int n, m;
int h[N], e[N], ne[N], w[N], idx;
int dist[N];
bool st[N];

void add(int a, int b, int c)
{
    e[idx] = b;
    w[idx] = c;
    ne[idx] = h[a];
    h[a] = idx ++;
}

int dijkstra()
{
    memset(dist, 0x3f, sizeof dist);
    dist[1] = 0;

    priority_queue<PII, vector<PII>, greater<PII>> heap;
    heap.push({0, 1});

    while (heap.size())
    {
        auto t = heap.top();
        heap.pop();

        int distance = t.first, ver = t.second;  // 读出第一个点，就是最短的dist值的那个点

        if (!st[ver])   // 实际上找的就是那个dist最小且st状态仍未被确定的点
        {
            st[ver] = true;
            for (int i = h[ver]; i != -1; i = ne[i])
            {
                int nd = e[i];

                if (dist[nd] > distance + w[i])
                {
                    dist[nd] = distance + w[i];
                    heap.push({dist[nd], nd});  //只有被更新了点，才添加到里面//当然直接添加所有的点也可以，冗余情况更多。
                }
            }
        }
    }

    return dist[n];
}


int main()
{
    scanf("%d%d", &n, &m);

    memset(h, -1, sizeof h);

    while (m --)
    {
        int a, b, c;
        scanf("%d%d%d", &a, &b, &c);
        add(a, b, c);
    }

    int t = dijkstra();

    if (t == 0x3f3f3f3f) cout << -1;
    else cout << t;

    return 0;
}