After a long party Petya decided to return home, but he turned out to be at the opposite end of the town from his home. There are n crossroads in the line in the town, and there is either the bus or the tram station at each crossroad.

The crossroads are represented as a string s of length n, where s_i = A, if there is a bus station at i-th crossroad, and s_i = B, if there is a tram station at i-th crossroad. Currently Petya is at the first crossroad (which corresponds to s_1) and his goal is to get to the last crossroad (which corresponds to s_n).

If for two crossroads i and j for all crossroads i, i+1, …, j-1 there is a bus station, one can pay a roubles for the bus ticket, and go from i-th crossroad to the j-th crossroad by the bus (it is not necessary to have a bus station at the j-th crossroad). Formally, paying a roubles Petya can go from i to j if s_t = A for all i ≤ t < j. 

If for two crossroads i and j for all crossroads i, i+1, …, j-1 there is a tram station, one can pay b roubles for the tram ticket, and go from i-th crossroad to the j-th crossroad by the tram (it is not necessary to have a tram station at the j-th crossroad). Formally, paying b roubles Petya can go from i to j if s_t = B for all i ≤ t < j.

For example, if s="AABBBAB", a=4 and b=3 then Petya needs:

<image>

  * buy one bus ticket to get from 1 to 3, 
  * buy one tram ticket to get from 3 to 6, 
  * buy one bus ticket to get from 6 to 7. 



Thus, in total he needs to spend 4+3+4=11 roubles. Please note that the type of the stop at the last crossroad (i.e. the character s_n) does not affect the final expense.

Now Petya is at the first crossroad, and he wants to get to the n-th crossroad. After the party he has left with p roubles. He's decided to go to some station on foot, and then go to home using only public transport.

Help him to choose the closest crossroad i to go on foot the first, so he has enough money to get from the i-th crossroad to the n-th, using only tram and bus tickets.

Input

Each test contains one or more test cases. The first line contains the number of test cases t (1 ≤ t ≤ 10^4).

The first line of each test case consists of three integers a, b, p (1 ≤ a, b, p ≤ 10^5) — the cost of bus ticket, the cost of tram ticket and the amount of money Petya has.

The second line of each test case consists of one string s, where s_i = A, if there is a bus station at i-th crossroad, and s_i = B, if there is a tram station at i-th crossroad (2 ≤ |s| ≤ 10^5).

It is guaranteed, that the sum of the length of strings s by all test cases in one test doesn't exceed 10^5.

Output

For each test case print one number — the minimal index i of a crossroad Petya should go on foot. The rest of the path (i.e. from i to n he should use public transport).

Example

Input


5
2 2 1
BB
1 1 1
AB
3 2 8
AABBBBAABB
5 3 4
BBBBB
2 1 1
ABABAB


Output


2
1
3
1
6
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using vi = vector<int>;
using unit = unsigned;
using vl = vector<ll>;
using ull = unsigned long long;
using vvi = vector<vi>;
using pii = pair<int, int>;
using vvl = vector<vl>;
using pll = pair<ll, ll>;
using vp = vector<pii>;
void _main();
int main() {
  cin.tie(0);
  ios::sync_with_stdio(false);
  _main();
}
template <class T>
string join(const vector<T> &v) {
  stringstream s;
  for (int i = (int)(0); i < (int)(((long long)(v).size())); ++i)
    s << ' ' << v[i];
  return s.str().substr(1);
}
template <class T>
istream &operator>>(istream &i, vector<T> &v) {
  for (auto &x : v) {
    i >> v;
  }
  return i;
}
template <class T>
ostream &operator<<(ostream &o, const vector<T> &v) {
  o << "[";
  for (auto &x : v) o << x << ",";
  o << "]";
  return o;
}
template <class T>
ostream &operator<<(ostream &o, const deque<T> &v) {
  o << "deq[";
  for (auto &x : v) o << x << ",";
  o << "]";
  return o;
}
template <class T>
ostream &operator<<(ostream &o, const set<T> &v) {
  o << "{";
  for (auto &x : v) o << x << ",";
  o << "}";
  return o;
}
template <class T>
ostream &operator<<(ostream &o, const unordered_set<T> &v) {
  o << "{";
  for (auto &x : v) o << x << ",";
  o << "}";
  return o;
}
template <class T>
ostream &operator<<(ostream &o, const multiset<T> &v) {
  o << "{";
  for (auto &x : v) o << x << ",";
  o << "}";
  return o;
}
template <class T>
ostream &operator<<(ostream &o, const unordered_multiset<T> &v) {
  o << "{";
  for (auto &x : v) o << x << ",";
  o << "}";
  return o;
}
template <class T1, class T2>
ostream &operator<<(ostream &o, const pair<T1, T2> &p) {
  o << "(" << p.first << "," << p.second << ")";
  return o;
}
template <class TK, class TV>
ostream &operator<<(ostream &o, const map<TK, TV> &m) {
  o << "{";
  for (auto &x : m) o << x.first << "=>" << x.second << ",";
  o << "}";
  return o;
}
template <class TK, class TV>
ostream &operator<<(ostream &o, const unordered_map<TK, TV> &m) {
  o << "{";
  for (auto &x : m) o << x.first << "=>" << x.second << ",";
  o << "}";
  return o;
}
template <class T>
void YES(T c) {
  if (c)
    cout << "YES" << endl;
  else
    cout << "NO" << endl;
}
template <class T>
void Yes(T c) {
  if (c)
    cout << "Yes" << endl;
  else
    cout << "No" << endl;
}
template <class T>
void POSS(T c) {
  if (c)
    cout << "POSSIBLE" << endl;
  else
    cout << "IMPOSSIBLE" << endl;
}
template <class T>
void Poss(T c) {
  if (c)
    cout << "Possible" << endl;
  else
    cout << "Impossible" << endl;
}
template <class T>
void chmax(T &a, const T &b) {
  if (a < b) a = b;
}
template <class T>
void chmin(T &a, const T &b) {
  if (b < a) a = b;
}
template <class T>
T gcd(T a, T b) {
  return b ? gcd(b, a % b) : a;
}
template <class T>
T lcm(T a, T b) {
  return a / gcd(a, b) * b;
}
const double EPS = 1e-10;
const double PI = acos(-1.0);
const int INF = 1001002003;
const ll LINF = 1001002003004005006LL;
const int dx[] = {-1, 0, 1, 0, -1, -1, 1, 1};
const int dy[] = {0, 1, 0, -1, -1, 1, -1, 1};
int solve() {
  int a, b, p;
  string s;
  cin >> a >> b >> p >> s;
  reverse((s).begin(), (s).end());
  int n = ((long long)(s).size());
  map<char, int> m{
      {'A', a},
      {'B', b},
  };
  int i = 1;
  char now = s[1];
  p -= m[now];
  if (p < 0) return n;
  while (i < n) {
    if (i + 1 < n && s[i + 1] == now) {
      ++i;
      continue;
    }
    now = s[i + 1];
    p -= m[now];
    if (p < 0) return n - i;
    ++i;
  }
  return 1;
}
void _main() {
  int t;
  cin >> t;
  vi a(t);
  for (int i = (int)(0); i < (int)(t); ++i) a[i] = solve();
  for (int i = (int)(0); i < (int)(t); ++i) cout << a[i] << endl;
}
