Problem:
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Similarity of Subtrees

Define the depth of a node in a rooted tree by applying the following rules recursively:

The depth of a root node is 0.

The depths of child nodes whose parents are with depth $d$ are $d + 1$.

Let $S(T, d)$ be the number of nodes of $T$ with depth $d$. Two rooted trees $T$ and $T'$ are similar if and only if $S(T, d)$ equals $S(T', d)$ for all non-negative integer $d$.

You are given a rooted tree $T$ with $N$ nodes. The nodes of $T$ are numbered from 1 to $N$. Node 1 is the root node of $T$. Let $T_i$ be the rooted subtree of $T$ whose root is node $i$. Your task is to write a program which calculates the number of pairs $(i, j)$ such that $T_i$ and $T_j$ are similar and $i < j$.

Input

The input consists of a single test case.

$N$

$a_1$ $b_1$

$a_2$ $b_2$

...

$a_{N-1}$ $b_{N-1}$

The first line contains an integer $N$ ($1 \leq N \leq 100,000$), which is the number of nodes in a tree. The following $N -1$ lines give information of branches: the $i$-th line of them contains $a_i$ and $b_i$, which indicates that a node $a_i$ is a parent of a node $b_i$. ($1 \leq a_i, b_i \leq N, a_i \ne b_i$) The root node is numbered by 1. It is guaranteed that a given graph is a rooted tree, i.e. there is exactly one parent for each node except the node 1, and the graph is connected.

Output

Print the number of the pairs $(x, y)$ of the nodes such that the subtree with the root $x$ and the subtree with the root $y$ are similar and $x < y$.

Sample Input 1

5
1 2
1 3
1 4
1 5

Output for the Sample Input 1

6

Sample Input 2

6
1 2
2 3
3 4
1 5
5 6

Output for the Sample Input 2

2

Sample Input 3

13
1 2
1 3
2 4
2 5
3 6
3 7
4 8
4 9
6 10
7 11
8 12
11 13

Output for the Sample Input 3

14

Solution in C++14 (judge verdict: Wrong Answer)

#include <bits/stdc++.h>
using namespace std;
#define repl(i,a,b) for(int i=(int)(a);i<(int)(b);i++)
#define rep(i,n) repl(i,0,n)
#define mp(a,b) make_pair((a),(b))
#define pb(a) push_back((a))
#define all(x) (x).begin(),(x).end()
#define uniq(x) sort(all(x)),(x).erase(unique(all(x)),end(x))
#define fi first
#define se second
#define dbg(...) _dbg(#__VA_ARGS__, __VA_ARGS__)
void _dbg(string){cout<<endl;}
template<class H,class... T> void _dbg(string s,H h,T... t){int l=s.find(',');cout<<s.substr(0,l)<<" = "<<h<<", ";_dbg(s.substr(l+1),t...);}
template<class T,class U> ostream& operator<<(ostream& o, const pair<T,U> &p){o<<"("<<p.fi<<","<<p.se<<")";return o;}
template<class T> ostream& operator<<(ostream& o, const vector<T> &v){o<<"[";for(T t:v){o<<t<<",";}o<<"]";return o;}

#define INF 1120000000
#define MOD 1000000007
#define DOM 1000000009
#define BASE 1000000021

using P = pair<int,pair<long,long>>;
// depth, hash1, hash2

vector<int> tree[100005];
map<pair<int,pair<long,long>>,int> memo;

P dfs(int v, int from){
  long h1 = 1, h2=1;
  int d = 1;
  for(auto to : tree[v]) if(to!=from){
    P p = dfs(to,v);
    d = max(d, p.fi+1);
    h1 += p.se.fi*MOD;
    h1 %= BASE;
    h2 += p.se.se*DOM;
    h2 %= BASE;
  }
  P p = mp(d,mp(h1,h2));
  memo[p]++;
  return p;
}

int main(){
  int n;
  cin>>n;
  rep(i,n-1){
    int a,b;
    cin>>a>>b;
    a--;b--;
    tree[a].pb(b);
    tree[b].pb(a);
  }

  dfs(0,-1);

  long ans = 0;
  for(auto &p : memo){
    ans += p.se * (p.se-1) / 2;
  }

  cout << ans << endl;

  return 0;
}