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;
  void printv(vector<long long int>& v) {
    for (auto x : v) {
      cout << x << " ";
    }
    cout << endl;
  }
  void printvv(vector<vector<long long int>>& vv) {
    for (auto x : vv) {
      for (auto y : x) {
        cout << y << " ";
      }
      cout << endl;
    }
  }
  void printVS(vector<string> v) {
    for (auto x : v) {
      cout << x << endl;
    }
  }
  int gcd(int a, int b) {
    int R;
    while ((a % b) > 0) {
      R = a % b;
      a = b;
      b = R;
    }
    return b;
  }
  long long int power(long long int x, long long int y, long long int p) {
    long long int res = 1;
    x = x % p;
    while (y > 0) {
      if (y & 1) res = (res * x) % p;
      y = y >> 1;
      x = (x * x) % p;
    }
    return res;
  }
  long long int modInverse(long long int n, long long int p) {
    return power(n, p - 2, p);
  }
  long long int nCrModPFermat(long long int n, long long int r, long long int p) {
    if (r == 0) return 1;
    long long int fac[10];
    fac[0] = 1;
    for (long long int i = 1; i <= n; i++) fac[i] = (fac[i - 1] * i) % p;
    return (fac[n] * modInverse(fac[r], p) % p * modInverse(fac[n - r], p) % p) %
           p;
  }
  long long int nPrModPFermat(long long int n, long long int r, long long int p) {
    if (r == 0) return 1;
    long long int fac[10];
    fac[0] = 1;
    for (long long int i = 1; i <= n; i++) fac[i] = (fac[i - 1] * i) % p;
    return (fac[n] * modInverse(fac[n - r], p) % p) % p;
  }
  void sieve(vector<long long int>& ans) {
    long long int n = 10000 + 5;
    bool prime[10000 + 5];
    for (long long int j = 0; j < n; j++) {
      prime[j] = true;
    }
    for (long long int p = 2; p * p < n; p++) {
      if (prime[p] == true) {
        for (int i = p * p; i < n; i += p) prime[i] = false;
      }
    }
    for (int p = 2; p < n; p++)
      if (prime[p]) ans.push_back(p);
  }
  long long int gcd(long long int a, long long int b) {
    if (b == 0) return a;
    return gcd(b, a % b);
  }
  long long lcm(long long int a, long long int b) { return (a / gcd(a, b)) * b; }
  long long int countSetBits(long long int x) {
    long long int ans = 0;
    while (x > 0) {
      if (x % 2 == 1) ans++;
      x = x / 2;
    }
    return ans;
  }
  void swap(long long int& x, long long int& y) {
    long long int temp = x;
    x = y;
    y = temp;
    return;
  }
  void swap(vector<long long int>& x, vector<long long int>& y) {
    vector<long long int> temp = x;
    x = y;
    y = temp;
    return;
  }
  bool first(pair<long long int, long long int> pr1,
             pair<long long int, long long int> pr2) {
    return pr1.first < pr2.first;
  }
  int solve() {
    long long int n;
    cin >> n;
    string second;
    cin >> second;
    for (long long int j = 0; j < n - 1; j++) {
      if (second[j] != second[j + 1]) {
        cout << j + 1 << " " << j + 2 << endl;
        return 0;
      }
    }
    cout << -1 << " " << -1 << endl;
    return 0;
  }
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    long long int t = 1;
    cin >> t;
    while (t--) {
      solve();
    }
    return 0;
  }
  