#include <bits/stdc++.h>
#define mem(s,t) memset(s,t,sizeof(s))
#define pb(s) push_back(s)
#define inf 0x3f3f3f3f
#define mn 100010
typedef long long ll;
using namespace std;

struct Node
{
    int l,r;
    ll c,w;
    bool operator <(const Node &x)const
    {
        if (x.w==w) return x.r>r;
        return x.w<w;
    };
}e[100010];

bool cmp(Node x,Node y)
{
    return x.l<y.l;
}

int n,m,last;
priority_queue<Node> q;

int main()
{
    int i,j,k;
    scanf("%d%d",&n,&m);
    for (i=1;i<=m;i++)
        scanf("%d%d%lld",&e[i].l,&e[i].r,&e[i].c);
    sort(e+1,e+m+1,cmp);
    while (!q.empty()) q.pop();
    Node p;
    p.l=1; p.r=1; p.w=0;
    q.push(p);
    last=0; j=1;
    while (!q.empty())
    {
        Node p=q.top();
       // cout<<p.l<<' '<<p.r<<' '<<p.w<<endl;
        q.pop();
        if (p.r<=last) continue;
        while (e[j].l<=p.r&&j<=m)
        {
            e[j].w=e[j].c+p.w;
            q.push(e[j]);
            j++;
        }
        last=p.r;
        if (last==n)
        {
            printf("%lld\n",p.w);
            return 0;
        }
    }
    printf("-1\n");
    return 0;
}


### 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 mem(s,t) memset(s,t,sizeof(s))
#define pb(s) push_back(s)
#define inf 0x3f3f3f3f
#define mn 100010
typedef long long ll;
using namespace std;

struct Node
{
    int l,r;
    ll c,w;
    bool operator <(const Node &x)const
    {
        if (x.w==w) return x.r>r;
        return x.w<w;
    };
}e[100010];

bool cmp(Node x,Node y)
{
    return x.l<y.l;
}

int n,m,last;
priority_queue<Node> q;

int main()
{
    int i,j,k;
    scanf("%d%d",&n,&m);
    for (i=1;i<=m;i++)
        scanf("%d%d%lld",&e[i].l,&e[i].r,&e[i].c);
    sort(e+1,e+m+1,cmp);
    while (!q.empty()) q.pop();
    Node p;
    p.l=1; p.r=1; p.w=0;
    q.push(p);
    last=0; j=1;
    while (!q.empty())
    {
        Node p=q.top();
       // cout<<p.l<<' '<<p.r<<' '<<p.w<<endl;
        q.pop();
        if (p.r<=last) continue;
        while (e[j].l<=p.r&&j<=m)
        {
            e[j].w=e[j].c+p.w;
            q.push(e[j]);
            j++;
        }
        last=p.r;
        if (last==n)
        {
            printf("%lld\n",p.w);
            return 0;
        }
    }
    printf("-1\n");
    return 0;
}


```