#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
#define rep(i,n) for(ll i = 0;i<((ll)(n));i++)
#define reg(i,a,b) for(ll i = ((ll)(a));i<=((ll)(b));i++)
#define irep(i,n) for(ll i = ((ll)(n)-1);i>=0;i--)
#define ireg(i,a,b) for(ll i = ((ll)(b));i>=((ll)(a));i--)
template<class T = int> using V = vector<T>;
template<class T = int> using VV = V< V<T> >;

/*
右端を更新できなければやる必要ないのでは？
*/

struct d{ll l,r,c;};
bool operator<(const d &l, const d &r){
	if(l.l<r.l)return true;
	if(l.l==r.l){
		if(l.c<r.c)return true;
		if(l.c==r.c){
			return l.r>r.r;
		}
	}
	return false;
}

ll n,m,l[100010],r[100010],c[100010];
d p[100010];
ll dp[100010];

int main(void){
	cin>>n>>m;
	rep(i,m)cin>>p[i].l>>p[i].r>>p[i].c;
	reg(i,1,n)dp[i]=1e18;
	dp[1]=0;
	sort(p,p+m);
	rep(i,m){
		ireg(j,p[i].l+1,p[i].r){
			if(dp[j]<=dp[p[i].l]+p[i].c)break;
			dp[j] = dp[p[i].l]+p[i].c;
		}
		// printf("%lld,%lld,%lld\n",p[i].l,p[i].r,p[i].c);
	}
	cout<<(dp[n]==1e18?-1:dp[n])<<endl;
	return 0;
}
### Prompt

Please provide a cpp coded solution to the problem described below:

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

### Response

```cpp
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
#define rep(i,n) for(ll i = 0;i<((ll)(n));i++)
#define reg(i,a,b) for(ll i = ((ll)(a));i<=((ll)(b));i++)
#define irep(i,n) for(ll i = ((ll)(n)-1);i>=0;i--)
#define ireg(i,a,b) for(ll i = ((ll)(b));i>=((ll)(a));i--)
template<class T = int> using V = vector<T>;
template<class T = int> using VV = V< V<T> >;

/*
右端を更新できなければやる必要ないのでは？
*/

struct d{ll l,r,c;};
bool operator<(const d &l, const d &r){
	if(l.l<r.l)return true;
	if(l.l==r.l){
		if(l.c<r.c)return true;
		if(l.c==r.c){
			return l.r>r.r;
		}
	}
	return false;
}

ll n,m,l[100010],r[100010],c[100010];
d p[100010];
ll dp[100010];

int main(void){
	cin>>n>>m;
	rep(i,m)cin>>p[i].l>>p[i].r>>p[i].c;
	reg(i,1,n)dp[i]=1e18;
	dp[1]=0;
	sort(p,p+m);
	rep(i,m){
		ireg(j,p[i].l+1,p[i].r){
			if(dp[j]<=dp[p[i].l]+p[i].c)break;
			dp[j] = dp[p[i].l]+p[i].c;
		}
		// printf("%lld,%lld,%lld\n",p[i].l,p[i].r,p[i].c);
	}
	cout<<(dp[n]==1e18?-1:dp[n])<<endl;
	return 0;
}
```