#include <queue>
#include <vector>
#include <iostream>
#include <algorithm>
using namespace std;
const long long inf = 1LL << 60;
struct edge {
	int to; long long cost;
};
struct state {
	int pos; long long cost;
};
bool operator<(const state& s1, const state& s2) {
	return s1.cost > s2.cost;
}
int main() {
	int N, M;
	cin >> N >> M;
	vector<vector<edge> > G(N);
	for (int i = 0; i < M; ++i) {
		int L, R, C;
		cin >> L >> R >> C; --L, --R;
		G[L].push_back(edge{ R, C });
	}
	for (int i = 1; i < N; ++i) {
		G[i].push_back(edge{ i - 1, 0 });
	}
	vector<long long> dist(N, inf);
	dist[0] = 0;
	priority_queue<state> que;
	que.push(state{ 0, 0 });
	vector<bool> vis(N);
	while (!que.empty()) {
		int pos = que.top().pos; que.pop();
		if (vis[pos]) continue;
		vis[pos] = true;
		for (edge e : G[pos]) {
			if (dist[e.to] > dist[pos] + e.cost) {
				dist[e.to] = dist[pos] + e.cost;
				que.push(state{ e.to, dist[e.to] });
			}
		}
	}
	cout << (dist[N - 1] == inf ? -1 : dist[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 <queue>
#include <vector>
#include <iostream>
#include <algorithm>
using namespace std;
const long long inf = 1LL << 60;
struct edge {
	int to; long long cost;
};
struct state {
	int pos; long long cost;
};
bool operator<(const state& s1, const state& s2) {
	return s1.cost > s2.cost;
}
int main() {
	int N, M;
	cin >> N >> M;
	vector<vector<edge> > G(N);
	for (int i = 0; i < M; ++i) {
		int L, R, C;
		cin >> L >> R >> C; --L, --R;
		G[L].push_back(edge{ R, C });
	}
	for (int i = 1; i < N; ++i) {
		G[i].push_back(edge{ i - 1, 0 });
	}
	vector<long long> dist(N, inf);
	dist[0] = 0;
	priority_queue<state> que;
	que.push(state{ 0, 0 });
	vector<bool> vis(N);
	while (!que.empty()) {
		int pos = que.top().pos; que.pop();
		if (vis[pos]) continue;
		vis[pos] = true;
		for (edge e : G[pos]) {
			if (dist[e.to] > dist[pos] + e.cost) {
				dist[e.to] = dist[pos] + e.cost;
				que.push(state{ e.to, dist[e.to] });
			}
		}
	}
	cout << (dist[N - 1] == inf ? -1 : dist[N - 1]) << endl;
	return 0;
}
```