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

Solution (in CPP):

Solution:
#include <iostream>
#include <utility>
#include <algorithm>
#include <queue>
#include <vector>
#define pii pair<int, int>
#define piii pair<pii, int>
#define pll pair<long long, long long>
#define ff first
#define ss second

using namespace std;

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

    int n, m; cin >> n >> m;
    piii inp[m]; for(auto &i : inp) {cin >> i.ff.ff >> i.ff.ss >> i.ss; --i.ff.ff;}
    sort(inp, inp + m);

    priority_queue<pll, vector<pll>, greater<pll>> Q;
    Q.push({0, 1});
    for(auto &i : inp)
    {
        while(!Q.empty() && Q.top().ss <= i.ff.ff) Q.pop();
        if(Q.empty()) {cout << -1; return 0;}
        Q.push({Q.top().ff + i.ss, i.ff.ss});
    }

    while(!Q.empty() && Q.top().ss < n) Q.pop();
    if(Q.empty()) {cout << -1; return 0;}
    cout << Q.top().ff;
    return 0;
}