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;
const int maxn = 1e5 + 10;
const int maxm = 1e6 + 10;
const int maxc = 26;
const int inf = 0x3f3f3f3f;
const long long mod = 1e9 + 7;
const long long INF = 0x3f3f3f3f3f3f3f3f;
const double pi = acos(-1.0);
const double eps = 1e-8;
inline int read() {
  int f = 1, x = 0;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = x * 10 + ch - 48;
    ch = getchar();
  }
  return f * x;
}
long long gcd(long long a, long long b) { return !b ? a : gcd(b, a % b); }
long long lcm(long long a, long long b) { return a / gcd(a, b) * b; }
long long poww(long long x, long long y, long long p) {
  long long ans = 1;
  while (y) {
    if (y & 1) ans = ans * x % p;
    x = x * x % p;
    y >>= 1;
  }
  return ans;
}
bool is_prime(long long x) {
  if (x == 1) return 0;
  for (long long i = 2; i * i <= x; i++) {
    if (x % i == 0) return 0;
  }
  return 1;
}
int t;
long long a, b, p;
string s;
vector<pair<int, int> > pos;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> t;
  while (t--) {
    pos.clear();
    cin >> a >> b >> p >> s;
    int ans = 1;
    int bus = 0, tram = 0;
    int now = s[0] - 'A';
    if (!now)
      bus++;
    else
      tram++;
    pos.push_back(make_pair(0, now));
    for (int i = 1; i < s.size(); i++) {
      int cur = s[i] - 'A';
      if (i == s.size() - 1) {
        pos.push_back(make_pair(i, -1));
        continue;
      }
      if (cur != now) {
        if (!cur)
          bus++;
        else
          tram++;
        now = cur;
        pos.push_back(make_pair(i, now));
      }
    }
    for (int i = 0; i < pos.size(); i++) {
      int id = pos[i].first, k = pos[i].second;
      if (a * bus + b * tram <= p) {
        cout << pos[i].first + 1 << endl;
        break;
      } else {
        if (!k)
          bus--;
        else
          tram--;
      }
    }
  }
}
