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>
using namespace std;

const int N = 100005;

int n, m, l, r, c;
vector<pair<int, int>> ve[N];
vector<long long> eve[N];
multiset<long long> se = {0};

int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr);
    cin >> n >> m;
    while (m--) {
        cin >> l >> r >> c;
        ve[l].push_back({r, c});
    }
    eve[2].push_back({0});
    for (int i = 1; i <= n; i++) {
        for (long long &v : eve[i]) {
            se.erase(se.find(v));
        }
        if (se.empty()) {
            return cout << -1, 0;
        }
        long long tmp = *se.begin();
        if (i == n) {
            return cout << tmp, 0;
        }
        for (pair<int, int> &v : ve[i]) {
            se.insert(tmp + v.second);
            eve[v.first + 1].push_back(tmp + v.second);
        }
    }
}