description: You are given a string s, consisting of n letters, each letter is either 'a' or 'b'. The letters in the string are numbered from 1 to n.
  
  s[l; r] is a continuous substring of letters from index l to r of the string inclusive. 
  
  A string is called balanced if the number of letters 'a' in it is equal to the number of letters 'b'. For example, strings "baba" and "aabbab" are balanced and strings "aaab" and "b" are not.
  
  Find any non-empty balanced substring s[l; r] of string s. Print its l and r (1 ≤ l ≤ r ≤ n). If there is no such substring, then print -1 -1.
  
  Input
  
  The first line contains a single integer t (1 ≤ t ≤ 1000) — the number of testcases.
  
  Then the descriptions of t testcases follow.
  
  The first line of the testcase contains a single integer n (1 ≤ n ≤ 50) — the length of the string.
  
  The second line of the testcase contains a string s, consisting of n letters, each letter is either 'a' or 'b'.
  
  Output
  
  For each testcase print two integers. If there exists a non-empty balanced substring s[l; r], then print l r (1 ≤ l ≤ r ≤ n). Otherwise, print -1 -1.
  
  Example
  
  Input
  
  
  4
  1
  a
  6
  abbaba
  6
  abbaba
  9
  babbabbaa
  
  
  Output
  
  
  -1 -1
  1 6
  3 6
  2 5
  
  Note
  
  In the first testcase there are no non-empty balanced subtrings.
  
  In the second and third testcases there are multiple balanced substrings, including the entire string "abbaba" and substring "baba".
solution: #include <bits/stdc++.h>
  using namespace std;
  const int64_t INF = 0x3f3f3f3f;
  const int64_t INFLL = 0x3f3f3f3f3f3f3f3f;
  const int64_t MOD = 1e9 + 7;
  const double EPS = 1e-9;
  const double PI = acos(-1);
  mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
  int64_t rand(int64_t x, int64_t y) {
    return uniform_int_distribution<int64_t>(x, y)(rng);
  }
  int64_t mul(int64_t a, int64_t b) {
    int64_t ret = (1ll * (a % MOD) * (b % MOD)) % MOD;
    return (ret + MOD) % MOD;
  }
  int64_t add(int64_t a, int64_t b) {
    int64_t ret = (1ll * (a % MOD) + (b % MOD)) % MOD;
    return (ret + MOD) % MOD;
  }
  int64_t pow_exp(int64_t n, int64_t p) {
    if (!p) return 1;
    if (p & 1) return mul(n, pow_exp(n, p - 1));
    int64_t tmp = pow_exp(n, p / 2);
    return mul(tmp, tmp);
  }
  const int64_t MAX = 2e5 + 10;
  const int64_t NMAX = 2e5 + 10;
  const int64_t MMAX = 2e5 + 10;
  const int64_t LOG_MAX = ceil(log2(double(NMAX)));
  const int64_t BLOCK = ceil(sqrt(double(NMAX)));
  int32_t main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    ;
    int64_t t;
    cin >> t;
    while (t--) {
      int64_t n;
      cin >> n;
      string str;
      cin >> str;
      bool found = false;
      for (int64_t i = 1; i < n; i++) {
        if (str[i - 1] != str[i]) {
          cout << i << ' ' << i + 1 << '\n';
          found = true;
          break;
        }
      }
      if (!found) {
        cout << "-1 -1" << '\n';
      }
    }
  }
  