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
#include <bits/stdc++.h>
#define MAX 131072
using namespace std;

pair<pair<int, int>, int> edge[MAX];
map<int, long long> val;

int main(void)
{
	int N, M;
	cin >> N >> M;
	val[1] = 0LL;
	val[N] = 1LL * 998244353 * 998244353;
	for (int i = 0; i < M; i++) cin >> edge[i].first.first >> edge[i].first.second >> edge[i].second;
	sort(edge, edge + M);
	for (auto e : edge)
	{
		if (!e.second) break;
		auto lr = val.lower_bound(e.first.first);
		auto ee = val.lower_bound(e.first.second);
		if (ee->second > lr->second + e.second)
		{
			val[e.first.second] = lr->second + e.second;
			ee = val.upper_bound(e.first.second - 1);
			while (ee->second >= lr->second + e.second && ee != lr)
			{
				auto eee = ee;
				ee--;
				if (eee->first != e.first.second) val.erase(eee);
			}
		}
	}
	if (val[N] < 1LL * 998244353 * 998244353) cout << val[N] << endl;
	else cout << -1 << endl;
	return 0;
}
