Problem: Oscolcovo city has a campus consisting of n student dormitories, n universities and n military offices. Initially, the i-th dormitory belongs to the i-th university and is assigned to the i-th military office.

Life goes on and the campus is continuously going through some changes. The changes can be of four types:

  1. University aj merges with university bj. After that all the dormitories that belonged to university bj are assigned to to university aj, and university bj disappears. 
  2. Military office cj merges with military office dj. After that all the dormitories that were assigned to military office dj, are assigned to military office cj, and military office dj disappears. 
  3. Students of university xj move in dormitories. Lets kxj is the number of dormitories that belong to this university at the time when the students move in. Then the number of students in each dormitory of university xj increases by kxj (note that the more dormitories belong to the university, the more students move in each dormitory of the university). 
  4. Military office number yj conducts raids on all the dormitories assigned to it and takes all students from there. 



Thus, at each moment of time each dormitory is assigned to exactly one university and one military office. Initially, all the dormitory are empty.

Your task is to process the changes that take place in the campus and answer the queries, how many people currently live in dormitory qj.

Input

The first line contains two integers, n and m (1 ≤ n, m ≤ 5·105) — the number of dormitories and the number of queries, respectively.

Next m lines contain the queries, each of them is given in one of the following formats: 

  * «U aj bj» — merging universities; 
  * «M cj dj» — merging military offices; 
  * «A xj» — students of university xj moving in the dormitories; 
  * «Z yj» — a raid in military office yj; 
  * «Q qj» — a query asking the number of people in dormitory qj. 

All the numbers in the queries are positive integers and do not exceed n. It is guaranteed that at the moment of the query the universities and military offices, that are present in the query, exist.

Output

In the i-th line print the answer to the i-th query asking the number of people in the dormitory.

Examples

Input

2 7
A 1
Q 1
U 1 2
A 1
Z 1
Q 1
Q 2


Output

1
0
2


Input

5 12
U 1 2
M 4 5
A 1
Q 1
A 3
A 4
Q 3
Q 4
Z 4
Q 4
A 5
Q 5


Output

2
1
1
0
1

Note

Consider the first sample test: 

  * In the first query university 1 owns only dormitory 1, so after the query dormitory 1 will have 1 student. 
  * After the third query university 1 owns dormitories 1 and 2. 
  * The fourth query increases by 2 the number of students living in dormitories 1 and 2 that belong to university number 1. After that 3 students live in the first dormitory and 2 students live in the second dormitory. 
  * At the fifth query the number of students living in dormitory 1, assigned to the military office 1, becomes zero. 

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
const int inf = 0x3f3f3f3f, Inf = 0x7fffffff;
const long long INF = 0x7fffffffffffffff;
const double eps = 1e-10;
unsigned int seed = 19260817;
const unsigned int _RAND_MAX_ = 4294967295u;
__inline__ __attribute__((always_inline)) unsigned int Rand() {
  return seed = seed * 998244353u + 1000000007u;
}
template <typename _Tp>
_Tp gcd(const _Tp &a, const _Tp &b) {
  return (!b) ? a : gcd(b, a % b);
}
template <typename _Tp>
__inline__ __attribute__((always_inline)) _Tp abs(const _Tp &a) {
  return a >= 0 ? a : -a;
}
template <typename _Tp>
__inline__ __attribute__((always_inline)) _Tp max(const _Tp &a, const _Tp &b) {
  return a < b ? b : a;
}
template <typename _Tp>
__inline__ __attribute__((always_inline)) _Tp min(const _Tp &a, const _Tp &b) {
  return a < b ? a : b;
}
template <typename _Tp>
__inline__ __attribute__((always_inline)) void chmax(_Tp &a, const _Tp &b) {
  (a < b) && (a = b);
}
template <typename _Tp>
__inline__ __attribute__((always_inline)) void chmin(_Tp &a, const _Tp &b) {
  (b < a) && (a = b);
}
template <typename _Tp>
__inline__ __attribute__((always_inline)) bool _cmp(const _Tp &a,
                                                    const _Tp &b) {
  return abs(a - b) <= eps;
}
template <typename _Tp>
__inline__ __attribute__((always_inline)) void read(_Tp &x) {
  register char ch(getchar());
  bool f(false);
  while (ch < 48 || ch > 57) f |= ch == 45, ch = getchar();
  x = ch & 15, ch = getchar();
  while (ch >= 48 && ch <= 57)
    x = (((x << 2) + x) << 1) + (ch & 15), ch = getchar();
  if (f) x = -x;
}
template <typename _Tp, typename... Args>
__inline__ __attribute__((always_inline)) void read(_Tp &t, Args &...args) {
  read(t);
  read(args...);
}
__inline__ __attribute__((always_inline)) int read_str(char *s) {
  register char ch(getchar());
  while (ch == ' ' || ch == '\r' || ch == '\n') ch = getchar();
  register char *tar = s;
  *tar = ch, ch = getchar();
  while (ch != ' ' && ch != '\r' && ch != '\n' && ch != EOF)
    *(++tar) = ch, ch = getchar();
  return tar - s + 1;
}
const int N = 1000005;
struct node {
  int x, y;
};
std::vector<node> v[N];
int opt[N];
int X[N], Y[N];
struct Union_set1 {
  int fa[N];
  int tag[N];
  int node_cnt;
  __inline__ __attribute__((always_inline)) Union_set1() {
    for (int i = 0; i < N; ++i) fa[i] = i;
    node_cnt = 500001;
  }
  int find(int x) {
    if (x == fa[x]) return x;
    int ans = find(fa[x]);
    if (ans != fa[x]) chmax(tag[x], tag[fa[x]]);
    fa[x] = ans;
    return ans;
  }
  int getval(int x) {
    find(x);
    int ans = tag[x];
    if (fa[x] != x) chmax(ans, tag[fa[x]]);
    return ans;
  }
  void merge(int x, int y) {
    x = find(x);
    y = find(y);
    int cur = ++node_cnt;
    fa[x] = fa[y] = cur;
  }
} S1;
struct Union_set2 {
  int fa[N];
  long long tag[N];
  int siz[N];
  int node_cnt;
  __inline__ __attribute__((always_inline)) Union_set2() {
    for (int i = 0; i < N; ++i) fa[i] = i;
    node_cnt = 500001;
  }
  int find(int x) {
    if (x == fa[x]) return x;
    int ans = find(fa[x]);
    if (ans != fa[x]) tag[x] += tag[fa[x]];
    fa[x] = ans;
    return ans;
  }
  long long getval(int x) {
    find(x);
    long long ans = tag[x];
    if (fa[x] != x) ans += tag[fa[x]];
    return ans;
  }
  void merge(int x, int y) {
    x = find(x);
    y = find(y);
    int cur = ++node_cnt;
    fa[x] = fa[y] = cur;
    siz[cur] = siz[x] + siz[y];
  }
} S2;
long long ans[N];
int main() {
  int n, m;
  read(n, m);
  char ch;
  int x, y;
  for (int i = 1; i <= m; ++i) {
    ch = getchar();
    while (ch != 'U' && ch != 'M' && ch != 'A' && ch != 'Z' && ch != 'Q')
      ch = getchar();
    if (ch == 'U') {
      read(X[i], Y[i]);
      opt[i] = 1;
    } else if (ch == 'M') {
      read(x, y);
      S1.merge(x, y);
    } else if (ch == 'A') {
      opt[i] = 2;
      read(X[i]);
    } else if (ch == 'Z') {
      read(x);
      chmax(S1.tag[S1.find(x)], i);
    } else {
      opt[i] = 3;
      read(X[i]);
      v[S1.getval(X[i])].push_back((node){X[i], i});
    }
  }
  for (int i = 1; i <= n; ++i) S2.siz[i] = 1;
  for (int i = 1; i <= m; ++i) {
    if (opt[i] == 1) {
      S2.merge(X[i], Y[i]);
    } else if (opt[i] == 2) {
      S2.tag[S2.find(X[i])] += S2.siz[S2.find(X[i])];
    } else if (opt[i] == 3) {
      ans[i] += S2.getval(X[i]);
    }
    for (auto it : v[i]) {
      ans[it.y] -= S2.getval(it.x);
    }
  }
  for (int i = 1; i <= m; ++i) {
    if (opt[i] == 3) {
      printf("%lld\n", ans[i]);
    }
  }
  return 0;
}