#include <stdio.h>
#include <stdlib.h>
using namespace std;

int n,i,j,k,u,v,c,root,val;
int son[100005],Next[200005],ed[200005],len[200005],tot;
int size[100005],heavy[100005];
bool vis[100005];
long long ans;

void dfs(int x)
{
	vis[x]=true;
	size[x]=1;
	for(int i=son[x];i;i=Next[i])
	if(!vis[ed[i]])
	{
		dfs(ed[i]);
		size[x]+=size[ed[i]];
		if(size[ed[i]]>heavy[x])heavy[x]=size[ed[i]];
	}
	if(n-size[x]>heavy[x])heavy[x]=n-size[x];
	if(heavy[x]<val)val=heavy[x],root=x;
	vis[x]=false;
}

void dfs2(int x,long long deep)
{
	vis[x]=true;
	size[x]=1;ans+=deep*2;
	for(int i=son[x];i;i=Next[i])
	if(!vis[ed[i]])
	{
		dfs2(ed[i],deep+len[i]);
		size[x]+=size[ed[i]];
	}
}

int main()
{
	scanf("%d",&n);
	for(i=1;i<n;++i)
	{
		scanf("%d%d%d",&u,&v,&c);
		++tot;Next[tot]=son[u];son[u]=tot;ed[tot]=v;len[tot]=c;
		++tot;Next[tot]=son[v];son[v]=tot;ed[tot]=u;len[tot]=c;
	}
	val=n;
	dfs(1);
	dfs2(root,0);
	val=1000000000;
	for(i=son[root];i;i=Next[i])
	if(size[ed[i]]==n/2&&n%2==0)
	{
		printf("%lld\n",ans-len[i]);
		return 0;
	}
	else
	{
		if(len[i]<val)val=len[i];
	}
	printf("%lld\n",ans-val);
}
### Prompt

Develop a solution in cpp to the problem described below:

There is a tree with N vertices, numbered 1 through N. The i-th edge in this tree connects Vertices A_i and B_i and has a length of C_i.

Joisino created a complete graph with N vertices. The length of the edge connecting Vertices u and v in this graph, is equal to the shortest distance between Vertices u and v in the tree above.

Joisino would like to know the length of the longest Hamiltonian path (see Notes) in this complete graph. Find the length of that path.

Constraints

* 2 \leq N \leq 10^5
* 1 \leq A_i < B_i \leq N
* The given graph is a tree.
* 1 \leq C_i \leq 10^8
* All input values are integers.

Input

Input is given from Standard Input in the following format:


N
A_1 B_1 C_1
A_2 B_2 C_2
:
A_{N-1} B_{N-1} C_{N-1}


Output

Print the length of the longest Hamiltonian path in the complete graph created by Joisino.

Examples

Input

5
1 2 5
3 4 7
2 3 3
2 5 2


Output

38


Input

8
2 8 8
1 5 1
4 8 2
2 5 4
3 8 6
6 8 9
2 7 12


Output

132

### Response

```cpp
#include <stdio.h>
#include <stdlib.h>
using namespace std;

int n,i,j,k,u,v,c,root,val;
int son[100005],Next[200005],ed[200005],len[200005],tot;
int size[100005],heavy[100005];
bool vis[100005];
long long ans;

void dfs(int x)
{
	vis[x]=true;
	size[x]=1;
	for(int i=son[x];i;i=Next[i])
	if(!vis[ed[i]])
	{
		dfs(ed[i]);
		size[x]+=size[ed[i]];
		if(size[ed[i]]>heavy[x])heavy[x]=size[ed[i]];
	}
	if(n-size[x]>heavy[x])heavy[x]=n-size[x];
	if(heavy[x]<val)val=heavy[x],root=x;
	vis[x]=false;
}

void dfs2(int x,long long deep)
{
	vis[x]=true;
	size[x]=1;ans+=deep*2;
	for(int i=son[x];i;i=Next[i])
	if(!vis[ed[i]])
	{
		dfs2(ed[i],deep+len[i]);
		size[x]+=size[ed[i]];
	}
}

int main()
{
	scanf("%d",&n);
	for(i=1;i<n;++i)
	{
		scanf("%d%d%d",&u,&v,&c);
		++tot;Next[tot]=son[u];son[u]=tot;ed[tot]=v;len[tot]=c;
		++tot;Next[tot]=son[v];son[v]=tot;ed[tot]=u;len[tot]=c;
	}
	val=n;
	dfs(1);
	dfs2(root,0);
	val=1000000000;
	for(i=son[root];i;i=Next[i])
	if(size[ed[i]]==n/2&&n%2==0)
	{
		printf("%lld\n",ans-len[i]);
		return 0;
	}
	else
	{
		if(len[i]<val)val=len[i];
	}
	printf("%lld\n",ans-val);
}
```