#include <bits/stdc++.h>
using namespace std;
#define rep(i,n) for(int i=0;i<(n);i++)
#define rep1(i,n) for(int i=1;i<=(n);i++)
#define all(x) (x).begin(),(x).end()
#define V vector
typedef V<int> vi;
typedef V<vi> vvi;
typedef long long ll;
typedef pair<ll, ll> P;
constexpr auto INF = INT_MAX >> 1;
constexpr auto LINF = 5000000000000000;
constexpr auto MOD = 1000000007;

int main() {

	int n, m;
	cin >> n >> m;

	V<V<P>> e(n, V<P>(0));
	rep(i, n - 1)e[i + 1].push_back(make_pair(0, i));
	
	rep(i, m) {
		int l, r;
		ll c;
		cin >> l >> r >> c;
		l--;r--;
		e[l].push_back(make_pair(c, r));
	}

	V<ll> dist(n, LINF);dist[0] = 0;
	priority_queue<P, V<P>, greater<P>> pq;
	pq.push(make_pair(0, 0));
	
	while (!pq.empty()) {
		ll d = pq.top().first;
		int v = pq.top().second;pq.pop();
		if (dist[v] < d)continue;
		rep(i, e[v].size()) {
			int vv = e[v][i].second;
			ll dd = e[v][i].first;
			if (dist[vv] <= d + dd)continue;
			dist[vv] = d + dd;
			pq.push(make_pair(dist[vv], vv));
		}
	}

	if (dist[n - 1] == LINF)cout << -1 << endl;
	else cout << dist[n - 1] << endl;

}
### Prompt

Your task is to create 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;
#define rep(i,n) for(int i=0;i<(n);i++)
#define rep1(i,n) for(int i=1;i<=(n);i++)
#define all(x) (x).begin(),(x).end()
#define V vector
typedef V<int> vi;
typedef V<vi> vvi;
typedef long long ll;
typedef pair<ll, ll> P;
constexpr auto INF = INT_MAX >> 1;
constexpr auto LINF = 5000000000000000;
constexpr auto MOD = 1000000007;

int main() {

	int n, m;
	cin >> n >> m;

	V<V<P>> e(n, V<P>(0));
	rep(i, n - 1)e[i + 1].push_back(make_pair(0, i));
	
	rep(i, m) {
		int l, r;
		ll c;
		cin >> l >> r >> c;
		l--;r--;
		e[l].push_back(make_pair(c, r));
	}

	V<ll> dist(n, LINF);dist[0] = 0;
	priority_queue<P, V<P>, greater<P>> pq;
	pq.push(make_pair(0, 0));
	
	while (!pq.empty()) {
		ll d = pq.top().first;
		int v = pq.top().second;pq.pop();
		if (dist[v] < d)continue;
		rep(i, e[v].size()) {
			int vv = e[v][i].second;
			ll dd = e[v][i].first;
			if (dist[vv] <= d + dd)continue;
			dist[vv] = d + dd;
			pq.push(make_pair(dist[vv], vv));
		}
	}

	if (dist[n - 1] == LINF)cout << -1 << endl;
	else cout << dist[n - 1] << endl;

}
```