#include<bits/stdc++.h>

using namespace std;

int main(void) {
    long n, m;
    cin >> n >> m;

    vector<vector<pair<long,long>>> G(n);

    for(int i=0; i<m; i++) {
        long l, r, c;
        cin >> l >> r >> c;
        l--, r--;
        G[l].push_back({r, c});
    }
    for(int i=0; i<n-1; i++)
        G[i+1].push_back({i, 0});

    priority_queue<pair<long,long>> q;
    q.push({0, 0});

    vector<long> dis(n, 1e15);
    while(!q.empty()) {
        auto t = q.top(); q.pop();
        long d = -t.first;
        long v = t.second;
        if(dis[v] <= d) continue;
        dis[v] = d;

        for(auto w: G[v]) {
            if(dis[w.first] <= d + w.second) continue;
            q.push({-(d + w.second), w.first});
        }
    }
    cout << (dis[n-1] == 1e15 ? -1 : dis[n-1]) << endl;
}

### Prompt

Construct a CPP code solution to the problem outlined:

We have N points numbered 1 to N arranged in a line in this order.

Takahashi decides to make an undirected graph, using these points as the vertices. In the beginning, the graph has no edge. Takahashi will do M operations to add edges in this graph. The i-th operation is as follows:

* The operation uses integers L_i and R_i between 1 and N (inclusive), and a positive integer C_i. For every pair of integers (s, t) such that L_i \leq s < t \leq R_i, add an edge of length C_i between Vertex s and Vertex t.



The integers L_1, ..., L_M, R_1, ..., R_M, C_1, ..., C_M are all given as input.

Takahashi wants to solve the shortest path problem in the final graph obtained. Find the length of the shortest path from Vertex 1 to Vertex N in the final graph.

Constraints

* 2 \leq N \leq 10^5
* 1 \leq M \leq 10^5
* 1 \leq L_i < R_i \leq N
* 1 \leq C_i \leq 10^9

Input

Input is given from Standard Input in the following format:


N M
L_1 R_1 C_1
:
L_M R_M C_M


Output

Print the length of the shortest path from Vertex 1 to Vertex N in the final graph. If there is no shortest path, print `-1` instead.

Examples

Input

4 3
1 3 2
2 4 3
1 4 6


Output

5


Input

4 2
1 2 1
3 4 2


Output

-1


Input

10 7
1 5 18
3 4 8
1 3 5
4 7 10
5 9 8
6 10 5
8 10 3


Output

28

### Response

```cpp
#include<bits/stdc++.h>

using namespace std;

int main(void) {
    long n, m;
    cin >> n >> m;

    vector<vector<pair<long,long>>> G(n);

    for(int i=0; i<m; i++) {
        long l, r, c;
        cin >> l >> r >> c;
        l--, r--;
        G[l].push_back({r, c});
    }
    for(int i=0; i<n-1; i++)
        G[i+1].push_back({i, 0});

    priority_queue<pair<long,long>> q;
    q.push({0, 0});

    vector<long> dis(n, 1e15);
    while(!q.empty()) {
        auto t = q.top(); q.pop();
        long d = -t.first;
        long v = t.second;
        if(dis[v] <= d) continue;
        dis[v] = d;

        for(auto w: G[v]) {
            if(dis[w.first] <= d + w.second) continue;
            q.push({-(d + w.second), w.first});
        }
    }
    cout << (dis[n-1] == 1e15 ? -1 : dis[n-1]) << endl;
}

```