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 MAXN = 1e5 + 10, inf = 1e18 + 10;

long long n, m;
long long l, r, c;
long long d[MAXN];
set <pair <long long, long long> > st;
vector <pair <long long, long long> > adj[MAXN];

void dijk() {
	for (int i = 2; i <= n; i++)
		d[i] = inf;
    d[1] = 0;
    
	st.insert({0, 1});

	while (st.size()) {
		int x = st.begin() -> second;
		st.erase(st.begin());
		for (auto y: adj[x])
			if (d[y.first] > d[x] + y.second) {
				st.erase({d[y.first], y.first});
				d[y.first] = d[x] + y.second;
				st.insert({d[y.first], y.first});
			}
	}
}

int main() {
    cin >> n >> m;
    for (int i = n; i > 1; i--)
        adj[i].push_back({i - 1, 0});
   
    for (int i = 0; i < m; i++) {
        cin >> l >> r >> c;
        adj[l].push_back({r, c});
    }
    dijk();
    if (d[n] == inf)
        cout << -1 << '\n';
    else
        cout << d[n] << '\n';
    return 0;
}