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>
#define MOD (ll)(1e9 + 7)
#define MAX (ll)(2e5 + 4)
#define INF (ll)(1e17)
using namespace std;
using ll =  long long;
using ld = long double;

vector<pair<ll, ll>> ve[MAX];

int main () {
	ll n, m;
	cin >> n >> m;
	for (ll i = 0; i < m; i++) {
		ll l, r, c;
		cin >> l >> r >> c;
		ve[l].push_back (make_pair (c, r));
	}
	ll ne = 1, dp[n + 1];
	for (int i = 0; i <= n; i++) {
		dp[i] = INF;			
	}
	dp[1] = 0;
	set <pair<ll, ll>> sp;
	ll co = 0;
	for (ll i = 1; i < n; i++) {
		for (auto x : ve[i]) {
			sp.insert (make_pair (dp[i] + x.first, x.second));
		}
		while (sp.size () && sp.begin () -> second <= i) {
			sp.erase (sp.begin ());
		}
		if (sp.size ()) {
			dp[i + 1] = min (dp[i + 1], sp.begin () -> first);
		}
		else {
			cout << "-1\n";
			return 0;
		}
	}
	cout << dp[n] << "\n";
}