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 <vector>
#include <map>
#include <set>
#include <queue>
#include <string>
#include <iomanip>
#include <algorithm>
#include <cmath>
#include <stdio.h>
using namespace std;
#define int long long
int MOD = 1000000007;
signed main() {
	cin.tie(0);
	ios::sync_with_stdio(false);
	int N, M;
	cin >> N >> M;
	vector<vector<int> > g(N);
	vector<vector<int> > c(N);
	for (int i = 1; i < N; i++) {
		g[i].push_back(i - 1);
		c[i].push_back(0);
	}
	int a, b, C;
	for (int i = 0; i < M; i++) {
		cin >> a >> b >> C;
		a--; b--;
		g[a].push_back(b);
		c[a].push_back(C);
	}

	priority_queue<pair<int, int> > pq;
	int INF = (int)1 << 60;
	vector<int> D(N, INF);
	D[0] = 0;
	pq.push(make_pair(0, 0));

	while (pq.size()) {
		a = pq.top().second;
		pq.pop();

		for (int i = 0; i < g[a].size();i++) {
			if (D[g[a][i]] > D[a] + c[a][i]) {
				D[g[a][i]] = D[a] + c[a][i];
				pq.emplace(-D[g[a][i]], g[a][i]);
			}
		}
	}
	int res = D[N - 1];
	if (res == INF)res = -1;
	cout << res << endl;
}