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 = 1e5 + 5;
vector<pair<long long , int> > st[N] , en[N];
long long dp[N];

int main() {
    int n , m;
    scanf("%d%d" , &n , &m);

    int s , e , c;
    while (m--) {
        scanf("%d%d%d" , &s , &e , &c);
        en[e].push_back({s , c});
    }

    // first : shortest path to n
    // second : node
    multiset<pair<long long , int> > info;
    for (int i = n ;i >= 1 ;i--) {
        if (i == n)
            dp[i] = 0;
        else
            dp[i] = info.size() ? info.begin()->first : -1;

        if (dp[i] == -1) {
            puts("-1");
            return 0;
        }

        for (auto &p : en[i]) {
            info.insert({p.second + dp[i] , i});
            st[p.first].push_back({p.second + dp[i] , i});
        }

        for (auto &c : st[i])
            info.erase(info.find(c));
    }

    printf("%lld" , dp[1]);
}