#include <bits/stdc++.h>
using namespace std;
const long long inf = 1e18;
int main() {
	ios::sync_with_stdio(false);
	cin.tie(nullptr);
	int n, m;
	cin >> n >> m;
	vector<vector<pair<int, int> > > G;
	G.resize(n);
	for(int i = 0; i < m; ++i) {
		int l, r, c;
		cin >> l >> r >> c;
		--r;
		--l;
		G[l].emplace_back(r, c);
	}
	for(int i = n - 1; i > 0; --i) {
		G[i].emplace_back(i - 1, 0);
	}
	priority_queue<pair<long long, int>, vector<pair<long long, int> >, greater<pair<long long, int> > > q;
	vector<long long> d;
	d.resize(n, inf);
	d[0] = 0;
	q.emplace(0, 0);
	while(!q.empty()) {
		auto o = q.top();
		int u = o.second;
		q.pop();
		if(o.first > d[u]) {
			continue;
		}
		for(auto o : G[u]) {
			int v = o.first;
			int w = o.second;
			if(d[v] > d[u] + w) {
				d[v] = d[u] + w;
				q.emplace(d[v], v);
			}
		}
	}
	if(d[n - 1] == inf) {
		cout << -1 << '\n';
	} else {
		cout << d[n - 1] << '\n';
	}
	return 0;
}
### 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>
using namespace std;
const long long inf = 1e18;
int main() {
	ios::sync_with_stdio(false);
	cin.tie(nullptr);
	int n, m;
	cin >> n >> m;
	vector<vector<pair<int, int> > > G;
	G.resize(n);
	for(int i = 0; i < m; ++i) {
		int l, r, c;
		cin >> l >> r >> c;
		--r;
		--l;
		G[l].emplace_back(r, c);
	}
	for(int i = n - 1; i > 0; --i) {
		G[i].emplace_back(i - 1, 0);
	}
	priority_queue<pair<long long, int>, vector<pair<long long, int> >, greater<pair<long long, int> > > q;
	vector<long long> d;
	d.resize(n, inf);
	d[0] = 0;
	q.emplace(0, 0);
	while(!q.empty()) {
		auto o = q.top();
		int u = o.second;
		q.pop();
		if(o.first > d[u]) {
			continue;
		}
		for(auto o : G[u]) {
			int v = o.first;
			int w = o.second;
			if(d[v] > d[u] + w) {
				d[v] = d[u] + w;
				q.emplace(d[v], v);
			}
		}
	}
	if(d[n - 1] == inf) {
		cout << -1 << '\n';
	} else {
		cout << d[n - 1] << '\n';
	}
	return 0;
}
```