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;
#define rep(i,a,b) for(int i=a;i<b;++i)
#define debug(x) cerr<<#x<<": "<<x<<endl
#define all(x) x.begin(), x.end()
using ll = long long;
int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    int n, m;
    cin >> n >> m;
    vector<int> l(m), r(m);
    vector<ll> c(m);
    rep(i,0,m) {
        cin >> l[i] >> r[i] >> c[i];
        --l[i];
        --r[i];
    }
    vector<vector<pair<int,ll>>> g(n);
    rep(i,1,n) g[i].emplace_back(i-1,0);
    rep(i,0,m) g[l[i]].emplace_back(r[i],c[i]);
    const ll INFTY = LLONG_MAX;
    vector<ll> dist(n,INFTY);
    using T = pair<ll,int>;
    priority_queue<T,vector<T>,greater<T>> pq;
    pq.emplace(0,0);
    while(!pq.empty()) {
        ll d;
        int u;
        tie(d,u) = pq.top();
        pq.pop();
        if(d<dist[u]) {
            dist[u] = d;
            for(const auto &e:g[u]) {
                pq.emplace(d+e.second, e.first);
            }
        }
    }
    ll ans = dist[n-1];
    if(ans==INFTY) ans = -1;
    cout<<ans<<endl;
}