#include <bits/stdc++.h>
using namespace std;

const int N = 100005;
const long long INF = 1e18;

int n, m;
int l[N], r[N], c[N];
vector <int> lef[N], rig[N];
long long smallest[N];
long long dis[N];

int main() {
    ios_base::sync_with_stdio(false); cin.tie(0);

    cin >> n >> m;
    for (int i = 1; i <= m; ++i) {
        cin >> l[i] >> r[i] >> c[i];
        lef[l[i]].push_back(i);
        rig[r[i]].push_back(i);
    }
    
    multiset <long long> ms;
    for (int i = 1; i <= n; ++i) {
        for (int &id : rig[i-1]) {
            ms.erase(ms.find(smallest[id]));
        }
        if (i != 1) {
            if (!ms.empty()) {
                dis[i] = *(ms.begin());
            } else {
                dis[i] = INF;
            }
        }
        for (int &id : lef[i]) {
            smallest[id] = min(INF, dis[i] + c[id]);
            ms.insert(smallest[id]);
        }
    }

    if (dis[n] >= INF) {
        cout << -1 << endl;
    } else {
        cout << dis[n] << endl;
    }
}
### Prompt

Develop a solution in cpp to the problem described below:

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;

const int N = 100005;
const long long INF = 1e18;

int n, m;
int l[N], r[N], c[N];
vector <int> lef[N], rig[N];
long long smallest[N];
long long dis[N];

int main() {
    ios_base::sync_with_stdio(false); cin.tie(0);

    cin >> n >> m;
    for (int i = 1; i <= m; ++i) {
        cin >> l[i] >> r[i] >> c[i];
        lef[l[i]].push_back(i);
        rig[r[i]].push_back(i);
    }
    
    multiset <long long> ms;
    for (int i = 1; i <= n; ++i) {
        for (int &id : rig[i-1]) {
            ms.erase(ms.find(smallest[id]));
        }
        if (i != 1) {
            if (!ms.empty()) {
                dis[i] = *(ms.begin());
            } else {
                dis[i] = INF;
            }
        }
        for (int &id : lef[i]) {
            smallest[id] = min(INF, dis[i] + c[id]);
            ms.insert(smallest[id]);
        }
    }

    if (dis[n] >= INF) {
        cout << -1 << endl;
    } else {
        cout << dis[n] << endl;
    }
}
```