#include <bits/stdc++.h>
#define mp make_pair
#define eb emplace_back
#define fi first
#define se second
#define int long long
using namespace std;
using cd = complex <double>;
 
typedef pair <int, long long> pii;
 
const long long INF = 1e18;
const int mod = 998244353;//1000000007;
const double Pi = acos(-1);
 
void Fastio()
{
	ios_base::sync_with_stdio(false);
	cin.tie(0);
	cout.tie(0);
}

int n, m;
int Dist[100005];
vector <pii> Adj[100005];
priority_queue <pair <int, int>, vector <pii>, greater <pii> > pq;

signed main()
{
	cin >> n >> m;
	Dist[1] = 0;
	for(int i = 2; i <= n; i++)
	{
		Dist[i] = INF;
		Adj[i].eb(mp(i - 1, 0));
	}
	for(int i = 1; i <= m; i++)
	{
		int u, v, w;
		cin >> u >> v >> w;
		Adj[u].eb(mp(v, w));
		Adj[v].eb(mp(u, w));
	}
	pq.push(mp(0, 1));
	while(pq.empty() == false)
	{
		int node = pq.top().se, d = pq.top().fi;
		pq.pop();
		if(d == Dist[node])
		{
			for(auto x : Adj[node])
			{
				if(Dist[x.fi] > x.se + Dist[node])
				{
					Dist[x.fi] = x.se + Dist[node];
					pq.push(mp(Dist[x.fi], x.fi));
				}
			}
		}
	}
	if(Dist[n] == INF)
	{
		Dist[n] = -1;
	}
	cout << Dist[n];
}
### Prompt

Your task is to create a CPP solution to the following 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

### Response

```cpp
#include <bits/stdc++.h>
#define mp make_pair
#define eb emplace_back
#define fi first
#define se second
#define int long long
using namespace std;
using cd = complex <double>;
 
typedef pair <int, long long> pii;
 
const long long INF = 1e18;
const int mod = 998244353;//1000000007;
const double Pi = acos(-1);
 
void Fastio()
{
	ios_base::sync_with_stdio(false);
	cin.tie(0);
	cout.tie(0);
}

int n, m;
int Dist[100005];
vector <pii> Adj[100005];
priority_queue <pair <int, int>, vector <pii>, greater <pii> > pq;

signed main()
{
	cin >> n >> m;
	Dist[1] = 0;
	for(int i = 2; i <= n; i++)
	{
		Dist[i] = INF;
		Adj[i].eb(mp(i - 1, 0));
	}
	for(int i = 1; i <= m; i++)
	{
		int u, v, w;
		cin >> u >> v >> w;
		Adj[u].eb(mp(v, w));
		Adj[v].eb(mp(u, w));
	}
	pq.push(mp(0, 1));
	while(pq.empty() == false)
	{
		int node = pq.top().se, d = pq.top().fi;
		pq.pop();
		if(d == Dist[node])
		{
			for(auto x : Adj[node])
			{
				if(Dist[x.fi] > x.se + Dist[node])
				{
					Dist[x.fi] = x.se + Dist[node];
					pq.push(mp(Dist[x.fi], x.fi));
				}
			}
		}
	}
	if(Dist[n] == INF)
	{
		Dist[n] = -1;
	}
	cout << Dist[n];
}
```