比较裸的点剖分，访问到每个重心时，记录一个b数组，
代表当前子树中长度为i的边的数量是b[i]，因为是3的倍数，
所以边长mod 3保存就行了，然后记录一个sum数组，代表
当前子树中一个子节点为根的子树中的情况（类似b），然后
用这两个数组不断的更新答案就行了。
/**************************************************************
Problem:
User: BLADEVIL
Language: Pascal
Result: Accepted
Time: ms
Memory: kb
****************************************************************/
//By BLADEVIL
var
n :longint;
pre, other, len :array[..] of longint;
last :array[..] of longint;
l, top :longint;
size, stack, yy :array[..] of longint;
ff :array[..] of boolean;
root :longint;
b, sum :array[..] of longint;
ans :longint;
function gcd(x,y:longint):longint;
begin
if x<y then exit(gcd(y,x)) else
if y= then exit(x) else exit(gcd(y,x mod y));
end;
procedure connect(x,y,z:longint);
begin
inc(l);
pre[l]:=last[x];
last[x]:=l;
other[l]:=y;
len[l]:=z;
end;
procedure dfs_size(x,fa:longint);
var
q, p :longint;
begin
size[x]:=;
inc(top);
stack[top]:=x;
q:=last[x];
while q<> do
begin
p:=other[q];
if (ff[q]) or (p=fa) then
begin
q:=pre[q];
continue;
end;
dfs_size(p,x);
inc(size[x],size[p]);
q:=pre[q];
end;
yy[x]:=fa;
end;
procedure getroot(u:longint);
var
ms, s, x, p, q :longint;
i :longint;
begin
top:=;
dfs_size(u,);
ms:=maxlongint;
for i:= to top do
begin
x:=stack[i];
s:=size[u]-size[x];
q:=last[x];
while q<> do
begin
p:=other[q];
if (ff[q]) or (p=yy[x]) then
begin
q:=pre[q];
continue;
end;
if size[p]>s then s:=size[p];
q:=pre[q];
end;
if s<ms then
begin
ms:=s;
root:=x;
end;
end;
end;
procedure dfs_value(x,fa,lz:longint);
var
q, p :longint;
begin
inc(sum[lz mod ]);
q:=last[x];
while q<> do
begin
p:=other[q];
if (p=fa) or (ff[q]) then
begin
q:=pre[q];
continue;
end;
dfs_value(p,x,lz+len[q]);
q:=pre[q];
end;
end;
procedure solve(u:longint);
var
i, q, p :longint;
begin
getroot(u);
if top= then exit;
fillchar(b,sizeof(b),);
b[]:=;
top:=;
q:=last[root];
while q<> do
begin
p:=other[q];
if ff[q] then
begin
q:=pre[q];
continue;
end;
fillchar(sum,sizeof(sum),);
dfs_value(p,root,len[q]);
for i:= to do ans:=ans+b[i]*sum[-i];
ans:=ans+sum[]*b[];
for i:= to do inc(b[i],sum[i]);
q:=pre[q];
end;
q:=last[root];
while q<> do
begin
p:=other[q];
if ff[q] then
begin
q:=pre[q];
continue;
end;
ff[q]:=true;
ff[q xor ]:=true;
solve(p);
q:=pre[q];
end;
end;
procedure main;
var
i :longint;
x, y, z :longint;
ans1, ans2 :longint;
g :longint;
begin
read(n);
l:=;
fillchar(b,sizeof(b),$ff);
b[]:=;
for i:= to n- do
begin
read(x,y,z);
z:=z mod ;
connect(x,y,z);
connect(y,x,z);
end;
ans:=;
solve();
ans1:=*ans+n; ans2:=n*n;
g:=gcd(ans1,ans2);
writeln(ans1 div g,'/',ans2 div g);
end;
begin
main;
end.