#include <bits/stdc++.h>
#define rep(i,n) for (int i = 0; i < (n); ++i)
using namespace std;
typedef long long ll;

const ll INF = 1e18;
struct edge{
	int to;
	int cost;
};

// sを始点とした全頂点の最短経路長（ダイクストラ）
// sからtに到達不可ならば d[t] = INF
vector<ll> dijkstra(int s, vector<vector<edge>> &to){
	// <最短距離、頂点番号>
	typedef pair<ll,int> P;
	
	priority_queue<P, vector<P>, greater<P>> que;
	vector<ll> d(to.size(), INF);
	
	d[s] = 0;
	que.push(P(0,s));
	
	while(!que.empty()){
		P p = que.top();
		que.pop();
		int v = p.second;
		if(d[v] < p.first) continue;
		
		for(edge e : to[v]){
			if(d[e.to] > d[v] + e.cost){
				d[e.to] = d[v] + e.cost;
				que.push(P(d[e.to], e.to));
			}
		}
	}
	
	return d;
}

int main(){
	
	int n,m;
	cin >> n >> m;
	
	vector<vector<edge>> to(n);
	rep(i,m){
		int l,r,c;
		cin >> l >> r >> c;
		--l; --r;
		to[l].push_back(edge{r,c});
		// for(int i = r; i > l; i--) to[i].push_back(edge{i-1,0});
	}
	
	rep(i,n-1) to[i+1].push_back(edge{i,0});
	
	vector<ll> d = dijkstra(0,to);
	
	if(d[n-1]==INF) cout << -1 << endl;
	else cout << d[n-1] << endl;
	return 0;
}

### Prompt

Your challenge is to write a Cpp solution to the following 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

### Response

```cpp
#include <bits/stdc++.h>
#define rep(i,n) for (int i = 0; i < (n); ++i)
using namespace std;
typedef long long ll;

const ll INF = 1e18;
struct edge{
	int to;
	int cost;
};

// sを始点とした全頂点の最短経路長（ダイクストラ）
// sからtに到達不可ならば d[t] = INF
vector<ll> dijkstra(int s, vector<vector<edge>> &to){
	// <最短距離、頂点番号>
	typedef pair<ll,int> P;
	
	priority_queue<P, vector<P>, greater<P>> que;
	vector<ll> d(to.size(), INF);
	
	d[s] = 0;
	que.push(P(0,s));
	
	while(!que.empty()){
		P p = que.top();
		que.pop();
		int v = p.second;
		if(d[v] < p.first) continue;
		
		for(edge e : to[v]){
			if(d[e.to] > d[v] + e.cost){
				d[e.to] = d[v] + e.cost;
				que.push(P(d[e.to], e.to));
			}
		}
	}
	
	return d;
}

int main(){
	
	int n,m;
	cin >> n >> m;
	
	vector<vector<edge>> to(n);
	rep(i,m){
		int l,r,c;
		cin >> l >> r >> c;
		--l; --r;
		to[l].push_back(edge{r,c});
		// for(int i = r; i > l; i--) to[i].push_back(edge{i-1,0});
	}
	
	rep(i,n-1) to[i+1].push_back(edge{i,0});
	
	vector<ll> d = dijkstra(0,to);
	
	if(d[n-1]==INF) cout << -1 << endl;
	else cout << d[n-1] << endl;
	return 0;
}

```