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

typedef long long ll;
typedef pair <ll, ll> pii;

const ll N = 1e5 + 5, mod = 1e15;

ll dist[N];
vector <pii> adj[N];
ll n, m;

void dij() {
	for (int i = 0; i < n; i++)	dist[i] = mod;
	set <pii> s;
	dist[0] = 0;
	s.insert({0, 0});
	while(!s.empty()) {
		int v = (*s.begin()).second;
		s.erase(s.begin());
		for (auto e : adj[v]) {
			int w = e.second, u = e.first;
			if(dist[u] > dist[v] + w) {
				s.erase({dist[u], u});
				dist[u] = dist[v] + w;
				s.insert({dist[u], u});
			}
		}
	}
}

int main() {
	ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
	cin >> n >> m;
	vector <pair <pii , int> > edge;
	for (int i = 0; i < n - 1; i++) {
		adj[i + 1].push_back({i, 0});
	}
	for (int i = 0; i < m; i++) {
		int v, u, w;
		cin >> v >> u >> w;
		v--, u--;
		adj[v].push_back({u, w});
		adj[u].push_back({v, w});
	}
	dij();
	if(dist[n - 1] == mod)	cout << -1 << "\n";
	else
		cout << dist[n - 1] << "\n";
}
### 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 <bits/stdc++.h>
using namespace std;

typedef long long ll;
typedef pair <ll, ll> pii;

const ll N = 1e5 + 5, mod = 1e15;

ll dist[N];
vector <pii> adj[N];
ll n, m;

void dij() {
	for (int i = 0; i < n; i++)	dist[i] = mod;
	set <pii> s;
	dist[0] = 0;
	s.insert({0, 0});
	while(!s.empty()) {
		int v = (*s.begin()).second;
		s.erase(s.begin());
		for (auto e : adj[v]) {
			int w = e.second, u = e.first;
			if(dist[u] > dist[v] + w) {
				s.erase({dist[u], u});
				dist[u] = dist[v] + w;
				s.insert({dist[u], u});
			}
		}
	}
}

int main() {
	ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
	cin >> n >> m;
	vector <pair <pii , int> > edge;
	for (int i = 0; i < n - 1; i++) {
		adj[i + 1].push_back({i, 0});
	}
	for (int i = 0; i < m; i++) {
		int v, u, w;
		cin >> v >> u >> w;
		v--, u--;
		adj[v].push_back({u, w});
		adj[u].push_back({v, w});
	}
	dij();
	if(dist[n - 1] == mod)	cout << -1 << "\n";
	else
		cout << dist[n - 1] << "\n";
}
```