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<cstdio>
#include<queue>
#include<utility>
#include<vector>
using namespace std;
typedef pair<long, int> P;

const int N_MAX = 1e5;
int n, m;
vector<pair<int, int>> v[N_MAX];
priority_queue<P, vector<P>, greater<P>> q;

int main() {
    scanf("%d %d", &n, &m);
    for (int i = 0; i < m; i++) {
        int l, r, c;
        scanf("%d %d %d", &l, &r, &c);
        v[l - 1].emplace_back(r - 1, c);
    }
    
    long ans = 1e18;
    q.emplace(0, 0);
    for (int i = 0; i < n - 1; i++) {
        for (int j = 0, e = v[i].size(); j < e; j++) {
            while (!q.empty() && q.top().second < i) q.pop();
            if (q.empty()) {
                printf("-1\n");
                return 0;
            }
            if (v[i][j].first == n - 1) {
                ans = min(ans, q.top().first + v[i][j].second);
            }
            else {
                q.emplace(q.top().first + v[i][j].second, v[i][j].first);
            }
        }
    }

    printf("%ld\n", (ans == 1e18) ? -1 : ans);
    return 0;
}