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 <algorithm>
#include <iostream>
#include <set>
#include <vector>

using namespace std;

int main() {
    int n, m;
    cin >> n >> m;

    vector<vector<long long>> in(n, vector<long long>(0)), out(n, vector<long long>(0));
    vector<long long> c(m);
    for (int i = 0; i < m; i++) {
        int l, r;
        cin >> l >> r >> c[i];
        in[l - 1].push_back(i);
        out[r - 1].push_back(i);
    }

    vector<long long> dist(n, 1LL << 48), dist_c(m, 1LL << 48);
    dist[0] = 0;

    multiset<long long> dist_set;

    for (int i = 0; i < n; i++) {
        if (dist_set.size() > 0) {
            dist[i] = min(dist[i], *dist_set.begin());
        }

        for (int j = 0; j < static_cast<int>(in[i].size()); j++) {
            dist_c[in[i][j]] = dist[i] + c[in[i][j]];
            dist_set.insert(dist_c[in[i][j]]);
        }

        for (int j = 0; j < static_cast<int>(out[i].size()); j++) {
            dist_set.erase(dist_set.find(dist_c[out[i][j]]));
        }
    }

    if (dist[n - 1] < (1LL << 48)) {
        cout << dist[n - 1] << endl;
    } else {
        cout << "-1" << endl;
    }
}