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;

const long long N = 2e5 + 5, M = 998244353;
int n, m, j;
long long dist[N];
pair<pair<int, int>, int> p[N];
priority_queue<pair<long long, int> > pq;

int main()
{
    ios::sync_with_stdio(0);
    cin.tie(0);

    cin >> n >> m;
    for (int i = 0; i < m; i++)
    {
        int u, v, c;
        cin >> u >> v >> c;
        p[i] = {{u, v}, c};
    }

    for (int i = 1; i <= n; i++)
    {
        dist[i] = LONG_LONG_MAX;
    }

    sort(p, p + m);

    pq.push({0, 1});
    dist[1] = 0;

    while (!pq.empty())
    {
        int u = pq.top().second;
        long long dis = -pq.top().first;
        pq.pop();

        while (j < m && p[j].first.first <= u)
        {
            if (dist[p[j].first.second] > dis + p[j].second)
            {
                dist[p[j].first.second] = dis + p[j].second;
                pq.push({-dist[p[j].first.second], p[j].first.second});
            }
            j++;
        }
    }

    cout << (dist[n] >= 1e17 ? -1 : dist[n]) << "\n";
}