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:
// daughter of 671D codeforces
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
const int N = 1e6 + 1;

struct line {
	int l, r, w;
	bool operator < (const line &o) {
		return r < o.r;
	}
} lie[N];

int n, m;
ll dp[N];
vector<int> cur[N];

int main() {
	cin.tie(0)->sync_with_stdio(0);
	cin >> n >> m;
	for (int i = 1, l, r, w; i <= m; i++) {
		cin >> l >> r >> w;
		lie[i] = {l, r, w};
		cur[l].push_back(i);
	}
	memset(dp, 0x3f3f3f3f, sizeof dp);
	set<pair<ll, int>> active;
	ll ans = LLONG_MAX;
	lie[0] = {1, 1, 0};
	active.insert({0, 0});
	for (int i = 1; i <= n; i++) {
		while (!active.empty() && lie[active.begin()->second].r < i) {
			active.erase(active.begin());
		}
		for (auto& id : cur[i]) {
			if (active.empty()) {
				cout << -1;
				return 0;
			}
			auto t = active.begin();
			dp[id] = min(dp[id], t->first + lie[id].w);
			if (lie[id].r == n) {
				ans = min(ans, dp[id]);
			}
		}
		for (auto& id : cur[i]) {
			active.insert({dp[id], id});
		}
	}
	if (ans == LLONG_MAX) cout << -1;
	else cout << ans;
}