// In The Name Of The Queen
#include<bits/stdc++.h>
using namespace std;
const int N = 1e5 + 10;
int n, q, L[N], R[N], C[N];
long long dp[N];
vector < int > V[N], U[N];
int main()
{
    scanf("%d%d", &n, &q);
    for (int i = 1; i <= q; i ++)
    {
        scanf("%d%d%d", &L[i], &R[i], &C[i]);
        V[L[i]].push_back(i);
        U[R[i]].push_back(i);
    }
    dp[1] = 0;
    set < pair < long long , int > > ST;
    for (int i = 1; i <= n; i ++)
    {
        if (i > 1 && !ST.size())
            return !printf("-1\n");
        if (i > 1)
            dp[i] = ST.begin()->first;
        for (int id : V[i])
            ST.insert({C[id] + dp[i], id});
        for (int id : U[i])
            ST.erase({C[id] + dp[L[id]], id});
    }
    printf("%lld\n", dp[n]);
    return 0;
}
### Prompt

Your challenge is to write 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
// In The Name Of The Queen
#include<bits/stdc++.h>
using namespace std;
const int N = 1e5 + 10;
int n, q, L[N], R[N], C[N];
long long dp[N];
vector < int > V[N], U[N];
int main()
{
    scanf("%d%d", &n, &q);
    for (int i = 1; i <= q; i ++)
    {
        scanf("%d%d%d", &L[i], &R[i], &C[i]);
        V[L[i]].push_back(i);
        U[R[i]].push_back(i);
    }
    dp[1] = 0;
    set < pair < long long , int > > ST;
    for (int i = 1; i <= n; i ++)
    {
        if (i > 1 && !ST.size())
            return !printf("-1\n");
        if (i > 1)
            dp[i] = ST.begin()->first;
        for (int id : V[i])
            ST.insert({C[id] + dp[i], id});
        for (int id : U[i])
            ST.erase({C[id] + dp[L[id]], id});
    }
    printf("%lld\n", dp[n]);
    return 0;
}
```