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

Solution (in CPP):

Solution:
#include<stdio.h>
#include<vector>
#include<algorithm>
#include<queue>
using namespace std;
typedef long long ll;
typedef pair<ll, ll>pii;
bool f[303030];
vector<pii>pat[303030];
int main()
{
	int num, way;
	scanf("%d%d", &num, &way);
	for (int i = 0; i < num - 1; i++)pat[i + 1].push_back(make_pair(i, 0));
	for (int i = 0; i < way; i++)
	{
		int za, zb, zc;
		scanf("%d%d%d", &za, &zb, &zc);
		za--, zb--;
		pat[za].push_back(make_pair(zb, zc));
	}
	priority_queue<pii>que;
	que.push(make_pair(0, 0));
	for (;;)
	{
		if (que.empty())
		{
			printf("-1\n");
			return 0;
		}
		pii z = que.top();
		que.pop();
		int v = z.second;
		if (f[v])continue;
		if (v == num - 1)
		{
			printf("%lld\n", -z.first);
			return 0;
		}
		f[v] = true;
		for (int i = 0; i < pat[v].size(); i++)que.push(make_pair(z.first - pat[v][i].second, pat[v][i].first));
	}
}