#include<bits/stdc++.h>
#define ll long long
#define pb push_back
#define pii pair<int,int>
#define mp make_pair
#define fi first
#define se second
using namespace std;
const int mod = 1e9 + 7;
const int maxn = 2e5 + 233;
priority_queue<pair<ll,int> > q;
vector<int> g[maxn],w[maxn];
bool vis[maxn];
ll d[maxn];
int main()
{
	int n,m;cin>>n>>m;
	for(int i=1;i<=n;i++)d[i] = 1LL<<60;
	for(int i=1;i<=m;i++){
		int l,r,c;scanf("%d%d%d",&l,&r,&c);
		g[l].pb(r);
		w[l].pb(c);
	}
	for(int i=2;i<=n;i++){
		g[i].pb(i-1);
		w[i].pb(0);
	}
	d[1] = 0;
	q.push(mp(0,1));
	while(!q.empty()){
		int x = q.top().second;q.pop();
		if(vis[x])continue;
		for(int i=0;i<g[x].size();i++){
			int y = g[x][i];
			if(d[y]>d[x]+w[x][i]){
				d[y] = d[x] + w[x][i];
				q.push(mp(-d[y],y));
			}
		}
	}
	if(d[n]==1LL<<60)cout<<-1<<endl;
	else cout << d[n] << endl;
}

### Prompt

Construct a CPP code solution to the problem outlined:

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>
#define ll long long
#define pb push_back
#define pii pair<int,int>
#define mp make_pair
#define fi first
#define se second
using namespace std;
const int mod = 1e9 + 7;
const int maxn = 2e5 + 233;
priority_queue<pair<ll,int> > q;
vector<int> g[maxn],w[maxn];
bool vis[maxn];
ll d[maxn];
int main()
{
	int n,m;cin>>n>>m;
	for(int i=1;i<=n;i++)d[i] = 1LL<<60;
	for(int i=1;i<=m;i++){
		int l,r,c;scanf("%d%d%d",&l,&r,&c);
		g[l].pb(r);
		w[l].pb(c);
	}
	for(int i=2;i<=n;i++){
		g[i].pb(i-1);
		w[i].pb(0);
	}
	d[1] = 0;
	q.push(mp(0,1));
	while(!q.empty()){
		int x = q.top().second;q.pop();
		if(vis[x])continue;
		for(int i=0;i<g[x].size();i++){
			int y = g[x][i];
			if(d[y]>d[x]+w[x][i]){
				d[y] = d[x] + w[x][i];
				q.push(mp(-d[y],y));
			}
		}
	}
	if(d[n]==1LL<<60)cout<<-1<<endl;
	else cout << d[n] << endl;
}

```