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 pb push_back
#define mp make_pair
#define ll long long
using namespace std;
const int N = 1e5 + 5;

int n, m;
vector <pair <ll, int> > s[N];

ll d[N];
priority_queue <pair <ll, int> > pq;

int main()
{
    ios_base::sync_with_stdio(false);
    cin.tie(NULL); cout.tie(NULL);

    int l, r, c;

    cin >> n >> m;
    for(int i = 1; i <= m; i++)
    {
        cin >> l >> r >> c;
        s[l].pb({c, r});
    }

    for(int i = 1; i <= n; i++)
        sort(s[i].begin(), s[i].end());

    int done_to = 1;
    for(auto t: s[1])
        pq.push({-t.first, t.second});
    d[n] = -1;
    while(!pq.empty())
    {
        pair <ll, int> t = pq.top();
        pq.pop();

        for(; done_to + 1 <= t.second;)
        {
            done_to++;
            d[done_to] = -t.first;
            for(auto p: s[done_to])
                pq.push({t.first - p.first, p.second});
        }
    }

    cout << d[n] << endl;

    return 0;
}