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;

const int maxn = 110000;
const long long inf = (long long)1e18 + 1e9;

long long C[maxn];
int N;

inline void update(int x, long long val) {
	for(; x; x -= x & -x) C[x] = min(C[x], val);
}

inline long long get(int x) {
	long long ret = inf;
	for(; x <= N; x += x & -x) ret = min(ret, C[x]);
	return ret;
}

struct Dist {
	int l, r, c;
	Dist(int l = 0, int r = 0, int c = 0) : l(l), r(r), c(c) {}
} d[maxn];

inline bool operator < (const Dist& lhs, const Dist& rhs) {
	return lhs.l < rhs.l;
}

int main() {
	ios::sync_with_stdio(0);
	cin.tie(0); cout.tie(0);
	int n, m; cin >> n >> m;
	N = n;
	for(int i = 1; i <= n; ++ i) C[i] = inf;
	update(1, 0);
	for(int i = 0; i < m; ++ i) {
		cin >> d[i].l >> d[i].r >> d[i].c;
	}
	sort(d, d + m);
	int j = 0;
	for(int i = 1; i <= n; ++ i) {
		for(; j < m && d[j].l == i; ++ j) {
			update(d[j].r, get(d[j].l) + d[j].c);
		}
	}
	long long ans = get(n);
	cout << (ans == inf ? -1 : ans) << "\n";
	return 0;
}