#include <bits/stdc++.h>

using namespace std;

constexpr int N = 100000;
constexpr int M = 100000;

set<pair<long long, int>> active;
vector<int> e_open[N + 1];
vector<int> e_close[N + 1];
int l[M + 1];
int r[M + 1];
long long c[M + 1];
long long dist[N + 1];
long long comp[N + 1];

int main() {
	int n, m, p, i, j;

	scanf("%d%d", &n, &m);

	for (i = 1; i <= m; i++) {
		scanf("%d%d%lld", l + i, r + i, c + i);
		e_open[l[i]].push_back(i);
		e_close[r[i]].push_back(i);
	}

	dist[1] = 0;

	// Pushing initial ranges.
	for (j = 0; j < e_open[1].size(); j++) {
		p = e_open[1][j];
		comp[p] = dist[1] + c[p];
		active.insert({comp[p], p});
	}

	for (i = 2; i <= n; i++) {
		if (active.empty()) {
			break;
		}

		dist[i] = active.begin()->first;

		for (j = 0; j < e_open[i].size(); j++) {
			p = e_open[i][j];
			comp[p] = dist[i] + c[p];
			active.insert({comp[p], p});
		}

		for (j = 0; j < e_close[i].size(); j++) {
			p = e_close[i][j];
			active.erase({comp[p], p});
		}
	}

	if (i <= n) {
		printf("-1\n");
	}
	else {
		printf("%lld\n", dist[n]);
	}

	return 0;
}
### Prompt

Create a solution in Cpp for 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;

constexpr int N = 100000;
constexpr int M = 100000;

set<pair<long long, int>> active;
vector<int> e_open[N + 1];
vector<int> e_close[N + 1];
int l[M + 1];
int r[M + 1];
long long c[M + 1];
long long dist[N + 1];
long long comp[N + 1];

int main() {
	int n, m, p, i, j;

	scanf("%d%d", &n, &m);

	for (i = 1; i <= m; i++) {
		scanf("%d%d%lld", l + i, r + i, c + i);
		e_open[l[i]].push_back(i);
		e_close[r[i]].push_back(i);
	}

	dist[1] = 0;

	// Pushing initial ranges.
	for (j = 0; j < e_open[1].size(); j++) {
		p = e_open[1][j];
		comp[p] = dist[1] + c[p];
		active.insert({comp[p], p});
	}

	for (i = 2; i <= n; i++) {
		if (active.empty()) {
			break;
		}

		dist[i] = active.begin()->first;

		for (j = 0; j < e_open[i].size(); j++) {
			p = e_open[i][j];
			comp[p] = dist[i] + c[p];
			active.insert({comp[p], p});
		}

		for (j = 0; j < e_close[i].size(); j++) {
			p = e_close[i][j];
			active.erase({comp[p], p});
		}
	}

	if (i <= n) {
		printf("-1\n");
	}
	else {
		printf("%lld\n", dist[n]);
	}

	return 0;
}
```