Problem:
Score : 600 points

Problem Statement

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.

Sample Input 1

4 3
1 3 2
2 4 3
1 4 6

Sample Output 1

5

We have an edge of length 2 between Vertex 1 and Vertex 2, and an edge of length 3 between Vertex 2 and Vertex 4, so there is a path of length 5 between Vertex 1 and Vertex 4.

Sample Input 2

4 2
1 2 1
3 4 2

Sample Output 2

-1

Sample Input 3

10 7
1 5 18
3 4 8
1 3 5
4 7 10
5 9 8
6 10 5
8 10 3

Sample Output 3

28

Solution in C++ (GCC 9.2.1) (judge verdict: Accepted)

#include<iostream>
#include<cstdio>
#include<cstring>
#include<vector>
#include<cmath>
using namespace std;
typedef long long ll;
#define N 100010
const ll inf=0x3f3f3f3f3f3f3f3fll;
inline int read(){
	int x=0,f=1;
	char c=getchar();
	while(c<'0'||c>'9'){
		if(c=='-')f=-1;
		c=getchar();
	}
	while(c>='0'&&c<='9'){
		x=(x<<1)+(x<<3)+c-'0';
		c=getchar();
	}
	return x*f;
}
int n,m;
ll dp[N],mn[N][19];
vector<pair<int,int> > vec[N];
void Insert(int x){
	mn[x][0]=dp[x];
	for(int i=1;i<=18;++i){
		if(x<=(1<<i-1))break;
		mn[x][i]=min(mn[x][i-1],mn[x-(1<<i-1)][i-1]);
	}
}
ll Query(int l,int r){
	int k=log2(r-l+1);
	return min(mn[r][k],mn[l+(1<<k)-1][k]);
}
int main(){
	n=read(),m=read();
	for(int i=1;i<=m;++i){
		int l=read(),r=read(),c=read();
		vec[r].emplace_back(l,c);
	}
	memset(mn,0x3f,sizeof(mn));
	memset(dp,0x3f,sizeof(dp));
	dp[1]=0;
	Insert(1);
	for(int i=2;i<=n;++i){
		for(auto x:vec[i]){
			dp[i]=min(dp[i],Query(x.first,i-1)+x.second);
		}
		Insert(i);
	}
	printf("%lld\n",dp[n]==inf?-1:dp[n]);
	return 0;
}