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 int long long
using namespace std;
using PII = pair<int, int>;
const int INF = 1LL<<60;
const int MOD = (int)1e9 + 7;

struct Edge {
    int to;
    int cost;
};

signed main() {
    int n, m;
    cin >> n >> m;
    vector<vector<Edge>> g(n);
    for (int i = 0; i < m; i++) {
        int l, r, c;
        cin >> l >> r >> c;
        g[l - 1].push_back(Edge{r - 1, c});
    }
    for (int i = 1; i < n; i++) {
        g[i].push_back(Edge{i - 1, 0});
    }

    priority_queue<PII, vector<PII>, greater<PII>> q;
    vector<int> d(n, INF);
    d[0] = 0;
    q.push({0, 0});
    while(!q.empty()) {
        PII p = q.top();
        q.pop();
        if (d[p.second] < p.first) continue;
        for (auto&& e : g[p.second]) {
            if (d[e.to] > d[p.second] + e.cost) {
                d[e.to] = d[p.second] + e.cost;
                q.push({d[e.to], e.to});
            }
        }
    }

    cout << (d[n - 1] == INF ? -1 : d[n - 1]) << "\n";
}