#include <bits/stdc++.h>
using namespace std;
long long int MOD = 1000000007;
long long int gcd(long long int a, long long int b) {
  if (b == 0)
    return a;
  else
    return gcd(b, a % b);
}
long long int lcm(long long int a, long long int b) {
  return a / gcd(a, b) * b;
}
long long int power(long long int a, long long int n) {
  if (n == 0) return 1;
  long long int p = power(a, n / 2);
  p = (p * p) % MOD;
  if (n % 2)
    return (p * a) % MOD;
  else
    return p % MOD;
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  long long int test = 1;
  cin >> test;
  while (test--) {
    long long int a, b, p;
    cin >> a >> b >> p;
    string s;
    cin >> s;
    long long int rem = p;
    long long int j = s.length() - 2;
    long long int pos = s.length();
    char prev = s[s.length() - 2];
    long long int count = 0;
    while (rem > 0 && j >= -1) {
      if (j == -1) {
        if (prev == 'A') {
          if (rem >= a) {
            rem -= a;
            pos -= count;
          } else {
            break;
          }
        } else {
          if (rem >= b) {
            rem -= b;
            pos -= count;
          } else {
            break;
          }
        }
        break;
      }
      if (s[j] == prev) {
        count++;
      } else {
        if (prev == 'A') {
          if (rem >= a) {
            rem -= a;
            pos -= count;
          } else {
            break;
          }
        } else {
          if (rem >= b) {
            rem -= b;
            pos -= count;
          } else {
            break;
          }
        }
        prev = s[j];
        count = 1;
      }
      j--;
    }
    cout << pos << '\n';
  }
  return 0;
}

### Prompt

Please create a solution in Cpp to 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;
long long int MOD = 1000000007;
long long int gcd(long long int a, long long int b) {
  if (b == 0)
    return a;
  else
    return gcd(b, a % b);
}
long long int lcm(long long int a, long long int b) {
  return a / gcd(a, b) * b;
}
long long int power(long long int a, long long int n) {
  if (n == 0) return 1;
  long long int p = power(a, n / 2);
  p = (p * p) % MOD;
  if (n % 2)
    return (p * a) % MOD;
  else
    return p % MOD;
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  long long int test = 1;
  cin >> test;
  while (test--) {
    long long int a, b, p;
    cin >> a >> b >> p;
    string s;
    cin >> s;
    long long int rem = p;
    long long int j = s.length() - 2;
    long long int pos = s.length();
    char prev = s[s.length() - 2];
    long long int count = 0;
    while (rem > 0 && j >= -1) {
      if (j == -1) {
        if (prev == 'A') {
          if (rem >= a) {
            rem -= a;
            pos -= count;
          } else {
            break;
          }
        } else {
          if (rem >= b) {
            rem -= b;
            pos -= count;
          } else {
            break;
          }
        }
        break;
      }
      if (s[j] == prev) {
        count++;
      } else {
        if (prev == 'A') {
          if (rem >= a) {
            rem -= a;
            pos -= count;
          } else {
            break;
          }
        } else {
          if (rem >= b) {
            rem -= b;
            pos -= count;
          } else {
            break;
          }
        }
        prev = s[j];
        count = 1;
      }
      j--;
    }
    cout << pos << '\n';
  }
  return 0;
}

```