Problem Description


Tina Town is a friendly place. People there care about each other.

A trie which was planted by the first mayor of Tina Town grows in the center of the town.

We define a palindrome substring in a trie a string that is a suffix of a string which the path from the root to any node represents and the string is a palindrome. Two palindromes are different if and only if their positions are different. Now, Tina wants to know the sum of the length of all palindrome substrings. Tina didn’t know the answer so she asked you to find out the answer for her.


 



Input


The first line contains a integer – number of cases
For each case, the first line is an integer  N representing the number of nodes in trie except the root.
The following  N lines contains a letter between  a and  d – the letter that node  n[i] stores and a number  f[i] – the index of  n[i]’s father. It’s guaranteed that  fa[i]≤i. If  fa[i]=0 n[i] is the root.
T≤10,N≤2∗106


 



Output


The first line contains a integer – number of cases
For each case, the first line is an integer  N representing the number of nodes in trie except the root.
The following  N lines contains a letter between  a and  d – the letter that node  n[i] stores and a number  f[i] – the index of  n[i]’s father. It’s guaranteed that  fa[i]≤i. If  fa[i]=0 n[i] is the root.
T≤10,N≤2∗106


 



Sample Input 


2
5
a 0
a 1
a 2
b 1
b 4
5
a 0
a 1
a 2
b 1
a 4

 



Sample Output 


14
15

Hint


回文树的应用，这里的操作有加入一个字母和删除一个字母，以及计算一个前缀对应的全部后缀回文串长度和，可以直接在回文树上统计。

#pragma comment(linker, "/STACK:102400000,102400000")
#include<map>
#include<set>
#include<cmath>
#include<queue>
#include<stack>
#include<bitset>
#include<cstdio>
#include<string>
#include<cstring>
#include<algorithm>
#include<functional>
using namespace std;
typedef long long LL;
const int low(int x) { return x&-x; }
const int INF = 0x7FFFFFFF;
const int mod = 1e9 + 7;
const int maxn = 2e6 + 10;
int T, n, ft[maxn], nt[maxn], x;
char s[maxn], ch;
LL ans;

struct PalindromicTree
{
  const static int maxn = 2e6 + 10;
  const static int size = 4;
  int next[maxn][size], sz, tot;
  int fail[maxn], len[maxn], last[maxn], pre[maxn];
  LL cnt[maxn];
  char s[maxn];
  void clear()
  {
    len[1] = -1; len[2] = 0;
    fail[1] = fail[2] = 1;
    cnt[1] = cnt[2] = tot = 0;
    last[0] = (sz = 3) - 1;
    memset(next[1], 0, sizeof(next[1]));
    memset(next[2], 0, sizeof(next[2]));
  }
  int Node(int length)
  {
    memset(next[sz], 0, sizeof(next[sz]));
    len[sz] = length;      return sz;
  }
  int getfail(int x)
  {
    while (s[tot] != s[tot - len[x] - 1]) x = fail[x];
    return x;
  }
  LL add(char pos)
  {
    int x = (s[++tot] = pos) - 'a', y = getfail(last[tot - 1]);
    if (next[y][x]) { last[tot] = next[y][x]; pre[tot] = 0; }
    else {
      last[tot] = next[y][x] = Node(len[y] + 2);
      pre[tot] = y;
      fail[sz] = len[sz] == 1 ? 2 : next[getfail(fail[y])][x];
      cnt[sz] = cnt[fail[sz]] + len[sz];        ++sz;
    }
    return cnt[last[tot]];
  }
  void del(char pos)
  {
    int x = pos - 'a';
    if (pre[tot])
    {
      next[pre[tot]][x] = 0;
      --sz;
    }
    --tot;
  }
}solve;

inline int get()
{
  while ((ch = getchar()) < '0' || ch > '9');
  int x = ch - '0';
  while ((ch = getchar()) >= '0' && ch <= '9') x = x * 10 + ch - '0';
  return x;
}

void dfs(int x)
{
  for (int i = ft[x]; i != -1; i = nt[i])
  {
    ans += solve.add(s[i]);
    dfs(i);
    solve.del(s[i]);
  }
}

int main()
{
  scanf("%d", &T);
  while (T--)
  {
    solve.clear();
    scanf("%d", &n);
    for (int i = 0; i <= n; i++) ft[i] = -1;
    for (int i = 1; i <= n; i++)
    {
      //while ((s[i] = getchar()) < 'a' || ch > 'd');
      //x = get(); nt[i] = ft[x];    ft[x] = i;
      scanf("%s%d", s + i, &x);
      nt[i] = ft[x];    ft[x] = i;
    }
    dfs(ans = 0);
    printf("%lld\n", ans);
  }
  return 0;
}