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;
#define ll long long


int main(){
    int N, M;
    cin >> N >> M;
    vector<vector<pair<int, ll>>> edge(N+1, vector<pair<int, ll>>());
    for (int i = 0; i < M; i++) {
        ll l, r, c;
        cin >> l >> r >> c;
        edge[l].push_back({r, c});
    }
    for (int i = 2; i <= N; i++) {
        edge[i].push_back({i-1, 0});
    }
    
    ll inf = LLONG_MAX;
    
    
    vector<ll> dist(N+1, inf);
    dist[1] = 0;
    priority_queue<pair<ll, int>, vector<pair<ll, int>>, greater<pair<ll, int>>> q;
    q.push({0, 1});
    while(!q.empty()) {
        int now = q.top().second;
        q.pop();
        for (auto p : edge[now]) {
            int nxt = p.first;
            ll c = p.second;
            if (dist[now] + c >= dist[nxt]) continue;
            dist[nxt] = dist[now] + c;
            q.push({dist[nxt], nxt});
        }
    }
    
    if (dist[N] == inf) cout << -1 << endl;
    else cout << dist[N] << endl;
    
    return 0;
}