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 N = 220000;

int L[N], R[N], id[N];
long long C[N], t[N];

void add(int u, long long w) {
	for (; u; u -= u & -u) t[u] = min(t[u], w);
}

long long get(int u) {
	long long ret = LLONG_MAX;
	for (; u < N; u += u & -u) ret = min(ret, t[u]);
	return ret;
}

int main() {
	ios::sync_with_stdio(0);
	cin.tie(0); cout.tie(0);
	int n, m; cin >> n >> m;
	for (int i = 0; i < m; i++) cin >> L[i] >> R[i] >> C[i];
	for (int i = 0; i < m; i++) id[i] = i;
	sort(id, id + m, [&](int u, int v) {
		if (L[u] != L[v]) return L[u] < L[v];
		return R[u] < R[v];
	});
	for (int i = 0; i < N; i++) t[i] = LLONG_MAX;
	add(1, 0);
	for (int i = 0; i < m; i++) {
		int u = id[i];
		int st = L[u], en = R[u]; long long w = C[u];
		long long now = get(st);
		if (now != LLONG_MAX) add(en, now + w);
	}
	long long ans = get(n);
	if (ans == LLONG_MAX) ans = -1;
	cout << ans << "\n";
	return 0;
}