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:
#ifdef LOCAL_EXEC
#include "include.h"
#else
#include <bits/stdc++.h>
using namespace std;
#define trace(...)
#endif

using LL = long long;
using PLI = pair<LL, int>;
const LL INF = 1e18;

int main() {
	ios_base::sync_with_stdio(false); cin.tie(NULL);

	int n, m; cin >> n >> m;
	vector<vector<int>> out(n);
	vector<vector<pair<int, int>>> in(n);
	for (int i = 0; i < m; i++) {
		int l, r, c; cin >> l >> r >> c; l--, r--;
		in[l].emplace_back(c, i);
		out[r].push_back(i);
	}

	vector<LL> dis(n, INF);
	priority_queue<PLI, vector<PLI>, greater<PLI>> heap;
	vector<bool> to_del(m);

	dis[0] = 0;
	for (int i = 0; i < n; i++) {
		while (!heap.empty() && to_del[heap.top().second]) heap.pop();
		if (!heap.empty()) {
			dis[i] = min(dis[i], heap.top().first);
		}

		for (auto& e: in[i]) {
			heap.emplace(dis[i] + e.first, e.second);
		}

		for (int x: out[i]) {
			to_del[x] = true;
		}
	}

	LL ans = dis[n - 1];
	if (ans == INF) ans = -1;
	cout << ans << endl;

	return 0;
}