#include <bits/stdc++.h>
using namespace std;
int main() {
  int t;
  cin >> t;
  for (int k = 0; k < t; k++) {
    int a, b, p;
    string s;
    cin >> a >> b >> p >> s;
    int price = 0;
    for (int i = 1; i <= s.size(); i++) {
      if (i == 1 || s[s.size() - 1 - i] != s[s.size() - i]) {
        if (s[s.size() - 1 - i] == 'A') {
          price += a;
        } else {
          price += b;
        }
        if (price > p) {
          cout << s.size() - i + 1 << endl;
          break;
        }
      }
      if (i == s.size()) cout << 1 << endl;
    }
  }
}

### Prompt

Create a solution in CPP for the following problem:

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

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
  int t;
  cin >> t;
  for (int k = 0; k < t; k++) {
    int a, b, p;
    string s;
    cin >> a >> b >> p >> s;
    int price = 0;
    for (int i = 1; i <= s.size(); i++) {
      if (i == 1 || s[s.size() - 1 - i] != s[s.size() - i]) {
        if (s[s.size() - 1 - i] == 'A') {
          price += a;
        } else {
          price += b;
        }
        if (price > p) {
          cout << s.size() - i + 1 << endl;
          break;
        }
      }
      if (i == s.size()) cout << 1 << endl;
    }
  }
}

```