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;
#define rep(i, srt, end) for (long long i = (srt); i < (long long)(end); i++)


int main(){
    ll N, M; cin >> N >> M;
    vector<vector<pair<ll,ll>>> G(N+1);
    rep(i, 0, M){
        ll v, u, w;
        cin >> v >> u >> w;
        G[u].push_back({v, w});
        G[v].push_back({u, w});
    }
    rep(i, 1, N) {
        G[i+1].push_back({i, 0});
    }

    ll srt = 1;
    constexpr ll INF = 1e18;
    vector<ll> d(N+1, INF);
    
    /* dijkstra's scope */{
        priority_queue<pair<ll,ll>, vector<pair<ll,ll>>, greater<pair<ll,ll>>> que;
        que.push({0, srt});
        d[srt] = 0;
        while(!que.empty()){
            auto q = que.top();
            que.pop();
            if(q.first > d[q.second]) continue;
            for(auto g : G[q.second]){
                if(d[g.first] <= q.first + g.second) continue;
                d[g.first] = q.first + g.second;
                que.push({d[g.first], g.first});
            }
        }
    }

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