Problem:
Problem F: Marked Ancestor

You are given a tree T that consists of N nodes. Each node is numbered from 1 to N, and node 1 is always the root node of T. Consider the following two operations on T:

M v: (Mark) Mark node v.

Q v: (Query) Print the index of the nearest marked ancestor of node v which is nearest to it. Initially, only the root node is marked. Note that a node is an ancestor of itself.

Your job is to write a program that performs a sequence of these operations on a given tree and calculates
the value that each Q operation will print. To avoid too large output file, your program is requested to
print the sum of the outputs of all query operations. Note that the judges confirmed that it is possible to
calculate every output of query operations in a given sequence.

Input

The input consists of multiple datasets. Each dataset has the following format:

The first line of the input contains two integers N and Q, which denotes the number of nodes in the tree
T and the number of operations, respectively. These numbers meet the following conditions: 1 ≤ N ≤ 100000 and 1 ≤ Q ≤ 100000.

The following N - 1 lines describe the configuration of the tree T. Each line contains a single integer pi (i = 2, ... , N), which represents the index of the parent of i-th node.

The next Q lines contain operations in order. Each operation is formatted as "M v" or "Q v", where v is the index of a node.

The last dataset is followed by a line containing two zeros. This line is not a part of any dataset and should not be processed.

Output

For each dataset, print the sum of the outputs of all query operations in one line.

Sample Input

6 3
1
1
2
3
3
Q 5
M 3
Q 5
0 0

Output for the Sample Input

4

Solution in C++14 (judge verdict: Accepted)

#include "bits/stdc++.h"
 
using namespace std;
typedef vector<int> VI;
typedef vector<VI> VVI;
typedef vector<VVI> VVVI;
typedef vector<string> VS;
typedef pair<int, int> PII;
typedef vector<PII> VPII;
typedef long long LL;
typedef priority_queue<int> PQ_DESC;
typedef priority_queue<int, vector<int>, greater<int>> PQ_ASC;
typedef priority_queue<PII> PQ_DESC_PII;
typedef priority_queue<PII, vector<PII>, greater<PII>> PQ_ASC_PII;
typedef vector<LL> VLL;
typedef vector<VLL> VVLL;
typedef vector<VVLL> VVVLL;

#define SORT_ASC(c) sort((c).begin(), (c).end())
#define SORT_DESC(c) sort((c).begin(), (c).end(), greater<typeof((c).begin())>())
 
#define FOR(i,a,b) for(int i=(a);i<(b);++i)
#define REP(i,n)  FOR(i,0,n)
#define FORL(i,a,b) for(LL i=(a);i<(b);++i)
#define REPL(i,n)  FORL(i,0,n)
#define SIZE(a) int((a).size())
#define ALL(a) (a).begin(),(a).end()
 
const double EPS = 1e-10;
const double PI  = acos(-1.0);

 //debug func
template<typename T>
void vprint(vector<T> v) {
    for(auto x : v) {
        cerr << x << " ";
    }
    cerr << endl;
}
 
template<typename T>
void vvprint(vector<vector<T>> vv) {
    REP(i, SIZE(vv)) {
        REP(j, SIZE(vv[i])) {
            cerr << vv[i][j] << " ";
        }
        cerr << endl;
    }
}

template <typename Iterator>
inline bool next_combination(const Iterator first, Iterator k, const Iterator last)
{
   /* Credits: Thomas Draper */
   if ((first == last) || (first == k) || (last == k))
      return false;
   Iterator itr1 = first;
   Iterator itr2 = last;
   ++itr1;
   if (last == itr1)
      return false;
   itr1 = last;
   --itr1;
   itr1 = k;
   --itr2;
   while (first != itr1)
   {
      if (*--itr1 < *itr2)
      {
         Iterator j = k;
         while (!(*itr1 < *j)) ++j;
         iter_swap(itr1,j);
         ++itr1;
         ++j;
         itr2 = k;
         rotate(itr1,j,last);
         while (last != j)
         {
            ++j;
            ++itr2;
         }
         rotate(k,itr2,last);
         return true;
      }
   }
   rotate(first,k,last);
   return false;
}

inline double get_time_sec(void){
    return static_cast<double>(chrono::duration_cast<chrono::nanoseconds>(chrono::steady_clock::now().time_since_epoch()).count())/1000000000;
}

template<typename T>
T gcd(T a, T b) {
    if(a > b) swap(a, b);
    if(a == 0) return b;
    else return gcd(b%a, a);
}

template<typename T>
T lcm(T a, T b) {
    return (a / gcd(a, b)) * b;
}

template<typename T>
map<T, T> prime_list(T n) {
    map<T, T> ret;
    for(T i = 2; i*i <= n; i++) {
        if(n % i == 0) {
            ret[i] = 0;
            while(n % i == 0) {
                n /= i;
                ret[i]++;
            }
        }
    }
    if(n != 1) ret[n]++;
    return ret;
}

#define MOD 1000000007

LL mypow(LL a, LL n) {
    if(n == 1) return a % MOD;
    if(n % 2 == 1) return (a * mypow(a, n-1)) % MOD;
    LL t = mypow(a, n/2);
    return (t * t) % MOD; 
}

LL mycomb(LL n, LL k) {
    if(n == 0 || n == k) return 1;
    LL x = 1, y = 1;
    REP(i, k) {
        x = x * (n-i) % MOD;
        y = y * (i+1) % MOD;
    }
    return x * mypow(y, MOD - 2) % MOD;
}

VI par, rnk, stp;

int root(int x) {
    if(stp[x]) return x;
    else return par[x] = root(par[x]);
}

bool same(int x, int y) {
    x = root(x);
    y = root(y);
    return x == y;
}

void unite(int x, int y) {
    x = root(x);
    y = root(y);
    if(rnk[x] < rnk[y]) {
        par[x] = y;
    } else {
        par[y] = x;
        if(rnk[x] == rnk[y]) rnk[x]++;
    }
}

#define INF 1e8

int main(void) {
    int n, q;
    while(1) {
        cin >> n >> q;
        if(n == 0 && q == 0) return 0;
        par = VI(n);
        iota(ALL(par), 0);
        rnk = VI(n, 1);
        stp = VI(n, 0);
        stp[0] = 1;
        FOR(i, 1, n) {
            int tmp;
            cin >> tmp;
            tmp--;
            par[i] = tmp;
        }
        LL ans = 0;
        VS s(q);
        VI v(q);
        for(int i = q-1; i >= 0; i--) {
            cin >> s[i] >> v[i];
            v[i]--;
            if(s[i] == "M") stp[v[i]]++; 
        }
        // vprint(par);
        // vprint(stp);
        REP(i, q) {
            if(s[i] == "M") {
                stp[v[i]]--;
            } else {
                ans += root(v[i]) + 1;
                // cerr << ans << endl;
            }
        }
        cout << ans << endl;
    }
}