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>
#pragma comment(linker, "/stack:200000000")
#pragma GCC optimize("Ofast", "omit-frame-pointer", "inline")
#pragma GCC option("arch=native", "tune=native", "no-zero-upper")
#pragma GCC target( \
    "sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native,avx2")
using namespace std;
const double eps = 1E-9;
const double Exp = 2.718281828459045;
const double Pi = 3.1415926535897932;
const int Max_Bit = 63;
const int INF = 2000000000;
const long long LINF = 1000000000000000007ll;
const int MOD = 1000000007;
const int NMAX = 1000005;
const int MMAX = 505;
const int KMAX = 6;
const string CODE = "CODEFORCES";
int t[4 * NMAX];
void build(vector<int> a, int v, int tl, int tr) {
  if (tl == tr)
    t[v] = a[tl];
  else {
    int tm = (tl + tr) / 2;
    build(a, v * 2, tl, tm);
    build(a, v * 2 + 1, tm + 1, tr);
    t[v] = t[v * 2] + t[v * 2 + 1];
  }
}
int sum(int v, int tl, int tr, int l, int r) {
  if (l > r) return 0;
  if (l == tl && r == tr) return t[v];
  int tm = (tl + tr) / 2;
  return sum(v * 2, tl, tm, l, min(r, tm)) +
         sum(v * 2 + 1, tm + 1, tr, max(l, tm + 1), r);
}
long long gcd(long long a, long long b, long long& x, long long& y) {
  if (a == 0) {
    x = 0;
    y = 1;
    return b;
  }
  long long x1, y1;
  long long d = gcd(b % a, a, x1, y1);
  x = y1 - (b / a) * x1;
  y = x1;
  return d;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int T;
  for (cin >> T; T; T--) {
    int a, b, p;
    cin >> a >> b >> p;
    string second;
    cin >> second;
    second[((int)(second).size()) - 1] = 'O';
    for (int i = ((int)(second).size()) - 1; i > 0; i--) {
      if (second[i - 1] == second[i]) {
        if (i == 1) cout << 1 << endl;
        continue;
      }
      if (second[i - 1] == 'A') {
        if (p < a) {
          cout << i + 1 << endl;
          break;
        } else {
          p -= a;
        }
      } else {
        if (p < b) {
          cout << i + 1 << endl;
          break;
        } else {
          p -= b;
        }
      }
      if (i == 1) cout << 1 << endl;
    }
  }
  return 0;
}
