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;

typedef pair<long long, long long> ii;
int n, m;
long long inf = (1ll << 61);
long long tree[200005];

void update(int i, long long v){
	i += n;
	tree[i] = v;
	while(i > 1){
		i /= 2;
		tree[i] = min(tree[i<<1],tree[i<<1|1]);
	}
}

long long query(int l, int r){
	long long ans = inf;
	for(l += n, r += n;l < r;l>>=1,r>>=1){
		if(l&1){
			ans = min(ans,tree[l]);
			l++;
		}
		if(r&1){
			r--;
			ans = min(ans,tree[r]);
		}
	}
	return ans;
}
int main(){
	//freopen("i.txt","r",stdin);
	ios_base::sync_with_stdio(false);
	cin >> n >> m;
	
	vector<ii> stuff[n];
	
	for(int i = 0;i < m;i++){
		int a, b, c;
		cin >> a >> b >> c;
		a--;
		b--;
		stuff[b].push_back(ii(a,c));
	}
	
	for(int i = 1;i < n;i++){
		update(i,inf);
	}
	
	update(0,0);
	
	for(int i = 0;i < n;i++){
		if(stuff[i].size() == 0) continue;
		long long ans = inf;
		for(ii x : stuff[i]){
			ans = min(ans, query(x.first, i) + x.second);
		}
		update(i,ans);
		//cout << i << " " << ans << "\n";
	}
	//cout << inf << "\n";
	long long ans = query(n-1,n);
	if(ans >= inf) cout << -1;
	else cout << ans;
	
}