#include <bits/stdc++.h>

using namespace std;
#define int long long
#define F first
#define S second
const int N = 1e5 + 10, inf = 1e18;
int dis[N];
vector < pair <int , int> > g[N];

void dijkstra(int v, int t)
{
    fill(dis, dis + N, inf);
    dis[v] = 0;
    priority_queue < pair <int , int> > pq;
    pq.push({0, v});
    while (pq.size()) {
        v = (pq.top()).S; pq.pop();
        if (v == t) return ;
        for (auto e : g[v]) {
            int u = e.F, w = e.S;
            if (dis[v] + w < dis[u]) {
                dis[u] = dis[v] + w;
                pq.push({-dis[u], u});
            }
        }
    }
}


int32_t main()
{
    int n, m; cin >> n >> m;
    for (int i = 2; i <= n; i ++) g[i].push_back({i - 1, 0});
    for (int i = 0; i < m; i ++) {
        int l, r, w; cin >> l >> r >> w;
        g[l].push_back({r, w});
    }
    dijkstra(1, n);
    cout << (dis[n] == inf ? -1 : dis[n]) << "\n";
}

### Prompt

Please create a solution in Cpp to the following problem:

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 int long long
#define F first
#define S second
const int N = 1e5 + 10, inf = 1e18;
int dis[N];
vector < pair <int , int> > g[N];

void dijkstra(int v, int t)
{
    fill(dis, dis + N, inf);
    dis[v] = 0;
    priority_queue < pair <int , int> > pq;
    pq.push({0, v});
    while (pq.size()) {
        v = (pq.top()).S; pq.pop();
        if (v == t) return ;
        for (auto e : g[v]) {
            int u = e.F, w = e.S;
            if (dis[v] + w < dis[u]) {
                dis[u] = dis[v] + w;
                pq.push({-dis[u], u});
            }
        }
    }
}


int32_t main()
{
    int n, m; cin >> n >> m;
    for (int i = 2; i <= n; i ++) g[i].push_back({i - 1, 0});
    for (int i = 0; i < m; i ++) {
        int l, r, w; cin >> l >> r >> w;
        g[l].push_back({r, w});
    }
    dijkstra(1, n);
    cout << (dis[n] == inf ? -1 : dis[n]) << "\n";
}

```