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 watch(x) cerr << (#x) << ": " << (x) << endl
#define int long long

const int N = 100009, inf = 1000000000000000000;
int n, m;

vector<pair<int, int>> graph[N];
int d[N];

void nope() {
    cout << -1 << endl;
    exit(0);
}

int32_t main() {

    ios_base::sync_with_stdio(false);
    cin.tie(NULL);

    cin >> n >> m;
    for(int i = 2; i <= n; ++i) {
        graph[i].emplace_back(i-1, 0);
    }
    for(int i = 0; i < m; ++i) {
        int l, r, c; cin >> l >> r >> c;
        graph[l].emplace_back(r, c);
    }

    for(int i = 1; i <= n; ++i) d[i] = inf;

    priority_queue<pair<int, int>> pq;
    pq.emplace(0, 1);

    while(!pq.empty()) {
        int w = -pq.top().first, v = pq.top().second; pq.pop();
        if(d[v] == inf) {
            d[v] = w;
            for(auto child : graph[v]) {
                if(d[child.first] == inf) {
                    pq.emplace(-w - child.second, child.first);
                }
            }
        }
    }

    if(d[n] == inf) cout << -1 << endl;
    else cout << d[n] << endl;
    
    return 0;
}