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 <iostream>
#include <vector>
#include <algorithm>
#include <queue>
using namespace std;
#define ll long long
#define rep(i,n) for(int i=0;i<(n);i++)
const ll INF = 1e15;

typedef pair<ll,ll> P;
struct edge{
    int to,cost;
};
int main(){
    cin.tie(0);
    ios::sync_with_stdio(false);
    int v,n;
    cin >> v >> n;
    ll d[v];
    vector<vector<edge>> G(v);
    rep(i,n){
        int a,b,c;
        cin >> a >> b >> c;
        a--; b--;
        G[a].push_back({b,c});
    }
    rep(i,v) if(i+1<v) G[i+1].push_back({i,0});
    priority_queue<P,vector<P>,greater<P>> pq;
    rep(i,v) d[i]=INF;
    d[0]=0;
    pq.push(P(0,0));
    while(!pq.empty()){
        P p=pq.top();
        pq.pop();
        int c = p.second;
        if(d[c]<p.first) continue;
        rep(i,G[c].size()){
            edge e = G[c][i];
            if(d[e.to] > d[c]+e.cost){
                d[e.to] = (ll)(d[c] + e.cost);
                pq.push({d[e.to],e.to});
            }
        }
    }
    
    if(d[v-1]<INF) cout << d[v-1] << endl;
    else cout << -1 << endl;
}