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;

#define MX 100050
#define INF 0x7f7f7f7f7f7f7f7fll
typedef long long ll;
struct edge {
	int v; ll w;
	edge(int v = 0, ll w = 0) : v(v), w(w) {}
	bool operator < (const edge& rhs) const { return w > rhs.w; }
};

vector<edge> g[MX];
ll d[MX];
bool vis[MX];

int main() {
	int n, m;
	scanf("%d%d", &n, &m);
	for(int i = 2; i <= n; i++) g[i].push_back(edge(i - 1, 0));
	while(m--) {
		int l, r, c;
		scanf("%d%d%d", &l, &r, &c);
		g[l].push_back(edge(r, c));
	}
	priority_queue<edge> pq;
	pq.push(edge(1, 0));
	for(int i = 2; i <= n; i++) d[i] = INF;
	
	while(pq.size()) {
		edge e = pq.top(); pq.pop();
		int u = e.v;
		if(vis[u]) continue;
		vis[u] = true;
		
		for(auto e : g[u])
			if(d[e.v] > d[u] + e.w) {
				d[e.v] = d[u] + e.w;
				pq.push(edge(e.v, d[e.v]));
			}
	}
	if(d[n] == INF) puts("-1");
	else printf("%lld\n", d[n]);
	return 0;
}