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 <bits/stdc++.h>
using namespace std;
 
//#pragma GCC optimize("O3")
//#pragma GCC target("sse4")
 
#define FOR(i, a, b) for(int i = (a); i < (b); ++i)
#define RFOR(i, b, a) for(int i = (b) - 1; i >= (a); --i)
#define FILL(a, value) memset(a, value, sizeof(a))
 
#define SZ(a) (int)((a).size())
#define ALL(a) a.begin(), a.end()
 
typedef long long LL;
typedef pair<int, int> PII;

const int INF = 1000 * 1000 * 1000 + 7;
const LL LINF = (LL)INF * INF;
const int MAX = 1 << 17;

int r[MAX], c[MAX];
vector<int> idxes[MAX];
vector<LL> vec[MAX];

int main()
{
	ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
	int n, m;
	cin >> n >> m;
	FOR(i, 0, m)
	{
		int l;
		cin >> l >> r[i] >> c[i];
		l--;
		idxes[l].push_back(i);
	}
	multiset<LL> s;
	FOR(i, 0, n)
	{
		for(LL x: vec[i])
			s.erase(s.find(x));
		if(i && s.empty())
		{
			cout << "-1\n";
			return 0;
		}
		LL d = i == 0 ? 0 : *s.begin();
		if(i == n - 1)
		{
			cout << d << "\n";
			return 0;
		}
		for(int id: idxes[i])
		{
			s.insert(d + c[id]);
			vec[r[id]].push_back(d + c[id]);
		}
	}
	assert(false);
	return 0;
}