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 <bits/stdc++.h>

#define long long long int
using namespace std;

// @author: pashka

int main() {
    ios::sync_with_stdio(false);

    int n, m;
    cin >> n >> m;
    vector<int> l(m), r(m), c(m);
    vector<vector<int>> beg(n);
    vector<vector<int>> end(n);
    for (int i = 0; i < m; i++) {
        cin >> l[i] >> r[i] >> c[i];
        l[i]--;
        r[i]--;
        beg[l[i]].push_back(i);
        end[r[i]].push_back(i);
    }
    vector<long> d(n, LONG_MAX);
    d[0] = 0;
    set<pair<long, int>> now;
    for (int i = 0; i < n; i++) {
        if (i > 0) {
            if (now.empty()) {
                cout << -1;
                return 0;
            }
            d[i] = now.begin() -> first;
//            cout << i << " " << d[i] << "\n";
        }
        for (int x : beg[i]) {
//            cout << i << " " << x << " " << d[l[x]] + c[x] << "\n";
            now.insert({d[l[x]] + c[x], x});
        }
        for (int x : end[i]) {
            now.erase({d[l[x]] + c[x], x});
        }
    }
    cout << d[n - 1] << "\n";

    return 0;
}