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 <algorithm>
#include <set>
#include <vector>
#include <map>

using namespace std;

int n, m;

struct Segm{
    int l, r;
    long long c;
};

Segm a[500000];

bool comp(Segm a, Segm b)
{
    return a.l < b.l;
}

vector<long long> d[500000];

int main()
{
    cin >> n >> m;
    for (int i = 0; i < m; i++) cin >> a[i].l >> a[i].r >> a[i].c;
    sort(a, a + m, comp);
    long long len = 0;
    map<long long, int> kek;
    int j = 0;
    for (int i = 1; i <= n; i++)
    {
        if (i > 1 && kek.size() == 0)
        {
            cout << -1;
            return 0;
        }
        if (i > 1)
        {
            auto it = kek.begin();
            len = (*it).first;
        }
        for (; j < m && a[j].l == i; j++)
        {
            kek[len + a[j].c]++;
            d[a[j].r].push_back(len + a[j].c);
        }
        for (int j = 0; j < d[i].size(); j++)
        {
            kek[d[i][j]]--;
            if (kek[d[i][j]] == 0) kek.erase(d[i][j]);
        }
    }
    cout << len;
}