#include <bits/stdc++.h>
#define rep(i,n) for(int i=(0);i<(n);i++)

using namespace std;

typedef long long ll;

// dijkstra 蟻本
struct edge {ll to, cost;};
typedef pair<ll, ll> P; // firstはsからの最短距離, secondは頂点の番号
const ll INF = LLONG_MAX / 2;
const int max_n = 101010;

int n;
vector<edge> g[max_n];
ll d[max_n];

void dijkstra(ll s){
	fill(d, d + n, INF);
	d[s] = 0;

	priority_queue<P,vector<P>,greater<P> > que;
	que.push(P(0,s));

	while(!que.empty()){
		P p = que.top();
		que.pop();

		ll v = p.second;
		if(d[v] < p.first) continue; //後から最短パスが見つかった場合、先にqueに入っているものはここではじかれる

		for(ll i = 0; i < g[v].size(); i++){
			edge e= g[v][i];
			if(d[e.to] > d[v] + e.cost){
				d[e.to] = d[v] + e.cost;
				que.push(P(d[e.to], e.to));
			}
		}
	}
}

int main()
{
	cin.tie(0);
	ios::sync_with_stdio(false);

	int m;
	cin >> n >> m;

	rep(i, m){
		int l, r;
		ll c;
		cin >> l >> r >> c;
		l--; r--;
		g[l].push_back((edge){r, c});
	}

  rep(i, n){
    g[i+1].push_back((edge){i, 0});
  }
  
  dijkstra(0);
  ll dist = d[n-1];

  if(dist == INF){
    cout << -1 << endl;
  }else{
    cout << dist << endl;
  }
	
}

### Prompt

Please formulate 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;

// dijkstra 蟻本
struct edge {ll to, cost;};
typedef pair<ll, ll> P; // firstはsからの最短距離, secondは頂点の番号
const ll INF = LLONG_MAX / 2;
const int max_n = 101010;

int n;
vector<edge> g[max_n];
ll d[max_n];

void dijkstra(ll s){
	fill(d, d + n, INF);
	d[s] = 0;

	priority_queue<P,vector<P>,greater<P> > que;
	que.push(P(0,s));

	while(!que.empty()){
		P p = que.top();
		que.pop();

		ll v = p.second;
		if(d[v] < p.first) continue; //後から最短パスが見つかった場合、先にqueに入っているものはここではじかれる

		for(ll i = 0; i < g[v].size(); i++){
			edge e= g[v][i];
			if(d[e.to] > d[v] + e.cost){
				d[e.to] = d[v] + e.cost;
				que.push(P(d[e.to], e.to));
			}
		}
	}
}

int main()
{
	cin.tie(0);
	ios::sync_with_stdio(false);

	int m;
	cin >> n >> m;

	rep(i, m){
		int l, r;
		ll c;
		cin >> l >> r >> c;
		l--; r--;
		g[l].push_back((edge){r, c});
	}

  rep(i, n){
    g[i+1].push_back((edge){i, 0});
  }
  
  dijkstra(0);
  ll dist = d[n-1];

  if(dist == INF){
    cout << -1 << endl;
  }else{
    cout << dist << endl;
  }
	
}

```