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;
typedef long long ll;
typedef tuple<int, ll, int> edges;

int main()
{
    int N, M;
    cin >> N >> M;
    set<edges> edgeSet;
    for(int i = 0 ; i < M; i ++){
        int L, R;
        ll C;
        cin >> L >> R >> C;
        edgeSet.insert(edges(L, C, R));
    }
    auto ite = edgeSet.begin();
    ll INF = LLONG_MAX;
    vector<ll> dp(N + 1, INF);
    dp.at(1) = 0;
    while (ite != edgeSet.end()) {  
        edges edge = *ite;
        int left =  get<0>(edge);
        ll cost =  get<1>(edge);
        int right = get<2>(edge);
        if(dp.at(left) == INF){
            ite++;
            continue;
        }
        ll currentDistance = cost + dp.at(left);
        for(int i = right; i >= left; i--){
            if(dp.at(i) <= currentDistance){
                break;
            }
            dp.at(i) = currentDistance;
        }
        ite++;
    }
    if(dp.at(N) == INF){
        cout << -1 << endl;
    }else{
        cout << dp.at(N) << endl;
    }
    return 0;
}