#include <bits/stdc++.h>
using namespace std;
#pragma GCC optimize("Ofast,no-stack-protector")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
#pragma GCC optimize("unroll-loops")
inline bool cmp(pair<long long, long long> a, pair<long long, long long> b) {
  if (a.first == b.first) return a.second < b.second;
  return a.first > b.first;
}
int drx[8] = {-2, -2, -1, -1, 1, 1, 2, 2};
int dcy[8] = {-1, 1, -2, 2, -2, 2, -1, 1};
int dirx[4] = {-1, 0, 1, 0};
int diry[4] = {0, -1, 0, 1};
template <class T>
inline T gcd(T a, T b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
template <class T>
inline T lcm(T a, T b) {
  return a * b / gcd<T>(a, b);
}
template <class T>
inline T power(T b, T p) {
  long long ans = 1;
  while (p--) ans *= b;
  return ans;
}
int main() {
  long long n, m, num, tc, t = 1;
  scanf("%lld", &tc);
  while (tc--) {
    scanf("%lld %lld", &n, &m);
    scanf("%lld", &num);
    string str;
    cin >> str;
    long long pos = str.length();
    str[pos - 1] = 'X';
    for (long long i = pos - 1; i > 0; --i) {
      if (str[i] != str[i - 1]) {
        if (str[i - 1] == 'A')
          num -= n;
        else
          num -= m;
      }
      if (num < 0) break;
      pos = i;
    }
    printf("%lld\n", pos);
  }
}

### Prompt

In Cpp, your task is to solve 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;
#pragma GCC optimize("Ofast,no-stack-protector")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
#pragma GCC optimize("unroll-loops")
inline bool cmp(pair<long long, long long> a, pair<long long, long long> b) {
  if (a.first == b.first) return a.second < b.second;
  return a.first > b.first;
}
int drx[8] = {-2, -2, -1, -1, 1, 1, 2, 2};
int dcy[8] = {-1, 1, -2, 2, -2, 2, -1, 1};
int dirx[4] = {-1, 0, 1, 0};
int diry[4] = {0, -1, 0, 1};
template <class T>
inline T gcd(T a, T b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
template <class T>
inline T lcm(T a, T b) {
  return a * b / gcd<T>(a, b);
}
template <class T>
inline T power(T b, T p) {
  long long ans = 1;
  while (p--) ans *= b;
  return ans;
}
int main() {
  long long n, m, num, tc, t = 1;
  scanf("%lld", &tc);
  while (tc--) {
    scanf("%lld %lld", &n, &m);
    scanf("%lld", &num);
    string str;
    cin >> str;
    long long pos = str.length();
    str[pos - 1] = 'X';
    for (long long i = pos - 1; i > 0; --i) {
      if (str[i] != str[i - 1]) {
        if (str[i - 1] == 'A')
          num -= n;
        else
          num -= m;
      }
      if (num < 0) break;
      pos = i;
    }
    printf("%lld\n", pos);
  }
}

```