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 <vector>
#include <set>
using namespace std;

#define ll long long
#define maxd 100000000000001

int main() {
    int n, m;
    cin >> n >> m;
    int l, r;
    ll c;
    vector<pair<int, ll>> adj[n + 1];
    set<pair<ll, int>> s;
    for (int i = 0; i < m; i++) {
        cin >> l >> r >> c;
        adj[l].emplace_back(r, c);
    }
    ll d[n + 1], temp;
    for (int i = 2; i <= n; i++) {
        d[i] = maxd;
        s.insert({maxd, i});
    }
    d[1] = 0;
    s.insert({0, 1});
    for (int i = 1; i < n; i++) {
        temp = s.begin()->first;
        for (auto p:adj[i]) {
            r = p.first;
            c = p.second;
            if (temp + c < d[r]) {
                s.erase({d[r], r});
                d[r] = temp + c;
                s.insert({d[r], r});
            }
        }
        s.erase({d[i], i});
    }
    if (d[n] == maxd) cout << "-1\n";
    else cout << d[n] << "\n";
    return 0;
}