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<queue>
#include<functional>

using namespace std;

typedef long long ll;
typedef pair<ll, ll> P;

int N, M;
vector<vector<P>> E;
ll d[100000];
const ll INF = 1145141919810893;

void djk() {
	for (int i = 0; i < N; i++)d[i] = INF;
	d[0] = 0;
	priority_queue<P, vector<P>, greater<P>> pq;
	pq.push(P(0, 0));
	while (pq.size()) {
		auto now = pq.top(); pq.pop();
		if (now.first > d[now.second])continue;
		for (int i = 0; i < E[now.second].size(); i++) {
			int nxt = E[now.second][i].first;
			ll cost = E[now.second][i].second;
			if (d[nxt] > d[now.second] + cost) {
				d[nxt] = d[now.second] + cost;
				pq.push(P(d[nxt], nxt));
			}
		}
	}
}

int main() {
	cin >> N >> M;
	E.resize(N);
	for (int i = 1; i < N; i++) {
		E[i].push_back(P(i - 1, 0));
	}
	for (int i = 0; i < M; i++) {
		int L, R;
		ll C;
		cin >> L >> R >> C;
		L--, R--;
		E[L].push_back(P(R, C));
	}

	djk();
	if (d[N - 1] == INF)cout << -1 << endl;
	else cout << d[N - 1] << endl;
	return 0;
}