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 <queue>
using namespace std;
typedef long long  ll;
typedef pair<ll,ll> pii;
const ll maxn = 1e6+10;
const ll inf = 1e14;

vector<pii> ve[maxn];
ll dis[maxn];
priority_queue<pii,vector<pii>,greater<pii> > heap;
int N,M;

void init(){
    for(int i = 1;i<=N;i++) dis[i] = inf;
    for(int i = 1;i<=N;i++) ve[i].push_back({i-1,0});
}

ll Dji(){
    heap.push({0,1});
    dis[1] = 0;
    
    while(heap.size()){
        auto cur = heap.top();heap.pop();
        ll d = cur.first,id = cur.second;
        for(auto v: ve[id]){
            ll id2 = v.first,w = v.second;
            if(d+w < dis[id2]){
                dis[id2] = d+w;
                heap.push({dis[id2],id2});
            }
        }
    }
    if(dis[N] == inf) return -1;
    else return dis[N];
}

int main(){
    cin>>N>>M;
    init();
    int a,b,c;
    for(int i = 0;i<M;i++){
        scanf("%d%d%d",&a,&b,&c);
        ve[a].push_back({b,c});
    }
    cout<<Dji()<<endl;
    
    return 0;
}