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

#define LL long long
const int N = 1e5 + 5;
int n, m, l, r, w;
vector<pair<LL, LL>> g[N];
LL d[N]; bool vis[N];

int main() {
    ios::sync_with_stdio(0), cin.tie(0);
    
    cin >> n >> m;
    while(m--) {
        cin >> l >> r >> w;
        g[l].push_back({r, w});
    }   
    for(int i = 1; i <= n; i++) d[i] = 4e18;
    for(int i = 2; i <= n; i++) g[i].push_back({i-1, 0});
    d[1] = 0;
    priority_queue<pair<LL, LL>, vector<pair<LL,LL>>, greater<pair<LL,LL>>> q;
    q.push({0, 1});

    while(!q.empty()) {
        pair<LL, LL> z = q.top();
        q.pop();
        if(z.second == n) {
            cout << z.first;
            return 0;
        }

        if(vis[z.second]) continue;
        vis[z.second] = 1;
        for(auto i : g[z.second]) {
            if(d[i.first] > z.first + i.second) {
                d[i.first] = z.first + i.second;
                q.push({d[i.first], i.first});  
            }
        }
    }
    cout << -1;
}
### 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;

#define LL long long
const int N = 1e5 + 5;
int n, m, l, r, w;
vector<pair<LL, LL>> g[N];
LL d[N]; bool vis[N];

int main() {
    ios::sync_with_stdio(0), cin.tie(0);
    
    cin >> n >> m;
    while(m--) {
        cin >> l >> r >> w;
        g[l].push_back({r, w});
    }   
    for(int i = 1; i <= n; i++) d[i] = 4e18;
    for(int i = 2; i <= n; i++) g[i].push_back({i-1, 0});
    d[1] = 0;
    priority_queue<pair<LL, LL>, vector<pair<LL,LL>>, greater<pair<LL,LL>>> q;
    q.push({0, 1});

    while(!q.empty()) {
        pair<LL, LL> z = q.top();
        q.pop();
        if(z.second == n) {
            cout << z.first;
            return 0;
        }

        if(vis[z.second]) continue;
        vis[z.second] = 1;
        for(auto i : g[z.second]) {
            if(d[i.first] > z.first + i.second) {
                d[i.first] = z.first + i.second;
                q.push({d[i.first], i.first});  
            }
        }
    }
    cout << -1;
}
```