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 = 1ll << 60ll;
ll n, m;
vector<pair<ll, ll> > g[100005];
ll min_cost[100005];

int main() {
	cin >> n >> m;
	
	for (int i = 1; i < n; i++) {
		g[i + 1].push_back({0, i});
	}
	
	for (int i = 1; i <= n; i++) {
		min_cost[i] = inf;
	}
	
	for (int i = 0; i < m; i++) {
		ll s, t, c;
		
		cin >> s >> t >> c;
		g[s].push_back({c, t});
	}
	
	priority_queue<pair<ll, ll>, vector<pair<ll, ll> >, greater<pair<ll, ll> > > p_que;
	
	p_que.push({0, 1});
	min_cost[1] = 0;
	while (!p_que.empty()) {
		ll pos = p_que.top().second;
		ll cos = p_que.top().first;
		p_que.pop();
		
		for (int i = 0; i < g[pos].size(); i++) {
			ll npos = g[pos][i].second;
			ll ncos = cos + g[pos][i].first;
			
			if (min_cost[npos] > ncos) {
				min_cost[npos] = ncos;
				p_que.push({ncos, npos});
			}
		}
	}
	
	if (min_cost[n] == inf) cout << -1 << endl;
	else cout << min_cost[n] << endl;
	
	return 0;
}