#include <bits/stdc++.h>
using namespace std;
#define ll long long
#define f(i, x, n) for (int i = x; i < (int)(n); ++i)

ll inf = 1e18;
int const N = 100000;
vector<pair<int, int> > ed[N + 1];

int main(){
	int n, m;
	scanf("%d%d", &n, &m);
	f(i, 0, m){
		int l, r, c;
		scanf("%d%d%d", &l, &r, &c);
		ed[r].emplace_back(l, c);
	}
	map<int, ll> st;
	st[1] = 0;
	f(i, 2, n + 1){
		ll an = inf;
		f(j, 0, ed[i].size()){
			int l = ed[i][j].first, c = ed[i][j].second;
			an = min(an, st.lower_bound(l)->second + c);
		}
		while (st.rbegin()->second >= an)st.erase(--st.end());
		st[i] = an;
	}
	printf("%lld\n", st.rbegin()->second >= inf ? -1 : st.rbegin()->second);
}
### Prompt

In CPP, your task is to solve 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;
#define ll long long
#define f(i, x, n) for (int i = x; i < (int)(n); ++i)

ll inf = 1e18;
int const N = 100000;
vector<pair<int, int> > ed[N + 1];

int main(){
	int n, m;
	scanf("%d%d", &n, &m);
	f(i, 0, m){
		int l, r, c;
		scanf("%d%d%d", &l, &r, &c);
		ed[r].emplace_back(l, c);
	}
	map<int, ll> st;
	st[1] = 0;
	f(i, 2, n + 1){
		ll an = inf;
		f(j, 0, ed[i].size()){
			int l = ed[i][j].first, c = ed[i][j].second;
			an = min(an, st.lower_bound(l)->second + c);
		}
		while (st.rbegin()->second >= an)st.erase(--st.end());
		st[i] = an;
	}
	printf("%lld\n", st.rbegin()->second >= inf ? -1 : st.rbegin()->second);
}
```