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>
#define rep(i,n) for (int i = 0; i < (n); ++i)
using namespace std;
typedef long long ll;

struct bb {
    ll a;
    ll b;
    ll c;
};

int main() {
    ll n, m;
    cin >> n >> m;
    vector<bb> imp;
    ll result[n+1];
    result[1] = 0;
    for(ll i=2;i<=n;i++) {
        result[i] = -1;
    }
    for(ll i=0; i < m;i++){
        ll a, b, c;
        cin >> a >> b >> c;
        bb sg = {a, b,c};
        imp.push_back(sg);
    }
    sort(imp.begin(), imp.end(), [](const bb& x, const bb& y){return x.a < y.a ||( x.a == y.a && (x.b < y.b || ( x.b == y.b && x.c < y.c))) ;} );
    for(vector<bb>::const_iterator i = imp.begin(); i != imp.end(); i++) {
        if (result[i->a] == -1) {
            cout << -1 << endl;
            return 0;
        }
        ll sh = result[i->a] + i->c;
        if (result[i->b] != -1 && result[i->b] <= sh) {
            continue;
        }
        for(ll j = i->a; j <= i->b; j++) {
            if (result[j] == -1 || result[j] > sh) {
                result[j] = sh;
            }
        }
    }
    cout << result[n] << endl;
}