#include <bits/stdc++.h>
using namespace std;

const int N = 110000;
const long long INF = 1LL << 60;

int L[N], R[N], C[N];
long long val[N];

vector<int> st[N], en[N];

set<pair<long long, int>> S;

int main() {
	
	ios::sync_with_stdio(0);
	cin.tie(0), cout.tie(0);
	
	int n, m; cin >> n >> m;
	for (int i = 1; i <= m; i++) {
		cin >> L[i] >> R[i] >> C[i];
		st[L[i]].push_back(i);
		en[R[i]].push_back(i);
	}
	
	for (int i = 1; i <= n; i++) val[i] = INF;
	val[1] = 0;
	for (int i = 1; i <= n; i++) {
		if (S.size()) val[i] = min(val[i], S.begin() -> first);
		for (auto it : st[i]) S.emplace(val[L[it]] + C[it], it);
		for (auto it : en[i]) S.erase(make_pair(val[L[it]] + C[it], it));
	}
	if (val[n] == INF) cout << "-1\n";
	else cout << val[n] << "\n";
	
	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;

const int N = 110000;
const long long INF = 1LL << 60;

int L[N], R[N], C[N];
long long val[N];

vector<int> st[N], en[N];

set<pair<long long, int>> S;

int main() {
	
	ios::sync_with_stdio(0);
	cin.tie(0), cout.tie(0);
	
	int n, m; cin >> n >> m;
	for (int i = 1; i <= m; i++) {
		cin >> L[i] >> R[i] >> C[i];
		st[L[i]].push_back(i);
		en[R[i]].push_back(i);
	}
	
	for (int i = 1; i <= n; i++) val[i] = INF;
	val[1] = 0;
	for (int i = 1; i <= n; i++) {
		if (S.size()) val[i] = min(val[i], S.begin() -> first);
		for (auto it : st[i]) S.emplace(val[L[it]] + C[it], it);
		for (auto it : en[i]) S.erase(make_pair(val[L[it]] + C[it], it));
	}
	if (val[n] == INF) cout << "-1\n";
	else cout << val[n] << "\n";
	
	return 0;
}

```