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;

using ll = long long;
const ll INF = 1e18;

const int MAXN = 1.1e5;
const int MAXM = 1.1e5;

int N, M, V;

vector<pair<int, int>> adj[MAXN];
ll dist[MAXN];

int main() {
	ios::sync_with_stdio(0), cin.tie(0);
	cin >> N >> M;
	for (int i = 0; i+1 < N; i++) {
		adj[i+1].emplace_back(i, 0);
	}
	for (int i = 0; i < M; i++) {
		int l, r, c; cin >> l >> r >> c; l--, r--;
		adj[l].emplace_back(r, c);
	}

	for (int i = 0; i < N; i++) {
		dist[i] = INF;
	}
	dist[0] = 0;
	priority_queue<pair<ll, int>, vector<pair<ll, int>>, greater<pair<ll, int>>> pq;
	pq.emplace(0, 0);
	while (!pq.empty()) {
		ll d; int cur; tie(d, cur) = pq.top();
		pq.pop();
		if (d != dist[cur]) continue;
		for (auto it : adj[cur]) {
			int nxt = it.first;
			ll nd = it.second + d;
			if (nd < dist[nxt]) {
				dist[nxt] = nd;
				pq.emplace(nd, nxt);
			}
		}
	}

	ll ans = dist[N-1];
	if (ans == INF) {
		ans = -1;
	}
	cout << ans << '\n';

	return 0;
}