#include <bits/stdc++.h>
using namespace std;
typedef long long int ll;
typedef pair<ll, ll> P;
typedef vector<ll> VI;
typedef vector<VI> VVI;
const ll MOD = 1000000007;
const ll INF = 1000000000000000;
#define REP(i, n) for(int i = 0; i < n; i++)
#define ALL(v) v.begin(), v.end()

struct edge { int to; int cost; };
vector<edge> G[200000];
vector<ll> d(200000, INF);
void dijkstra(int s) {
	priority_queue<P, vector<P>, greater<P> > que;
	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;
		REP(i,G[v].size()) {
			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() {
	int n, m; cin >> n >> m;
	REP(i, n) G[i + 1].push_back({ i,0 });
	int a, b, c;
	REP(i, m) {
		cin >> a >> b >> c;
		G[a].push_back({ b,c });
	}
	dijkstra(1);
	if (d[n] == INF)
		cout << -1;
	else
		cout << d[n] << endl;
	return 0;
}
### Prompt

Generate 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>
using namespace std;
typedef long long int ll;
typedef pair<ll, ll> P;
typedef vector<ll> VI;
typedef vector<VI> VVI;
const ll MOD = 1000000007;
const ll INF = 1000000000000000;
#define REP(i, n) for(int i = 0; i < n; i++)
#define ALL(v) v.begin(), v.end()

struct edge { int to; int cost; };
vector<edge> G[200000];
vector<ll> d(200000, INF);
void dijkstra(int s) {
	priority_queue<P, vector<P>, greater<P> > que;
	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;
		REP(i,G[v].size()) {
			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() {
	int n, m; cin >> n >> m;
	REP(i, n) G[i + 1].push_back({ i,0 });
	int a, b, c;
	REP(i, m) {
		cin >> a >> b >> c;
		G[a].push_back({ b,c });
	}
	dijkstra(1);
	if (d[n] == INF)
		cout << -1;
	else
		cout << d[n] << endl;
	return 0;
}
```