description: You are given two integers l and r, l≤ r. Find the largest possible value of a mod b over all pairs (a, b) of integers for which r≥ a ≥ b ≥ l.
  
  As a reminder, a mod b is a remainder we get when dividing a by b. For example, 26 mod 8 = 2.
  
  Input
  
  Each test contains multiple test cases.
  
  The first line contains one positive integer t (1≤ t≤ 10^4), denoting the number of test cases. Description of the test cases follows.
  
  The only line of each test case contains two integers l, r (1≤ l ≤ r ≤ 10^9).
  
  Output
  
  For every test case, output the largest possible value of a mod b over all pairs (a, b) of integers for which r≥ a ≥ b ≥ l.
  
  Example
  
  Input
  
  
  4
  1 1
  999999999 1000000000
  8 26
  1 999999999
  
  
  Output
  
  
  0
  1
  12
  499999999
  
  Note
  
  In the first test case, the only allowed pair is (a, b) = (1, 1), for which a mod b = 1 mod 1 = 0.
  
  In the second test case, the optimal choice is pair (a, b) = (1000000000, 999999999), for which a mod b = 1.
solution: #include <bits/stdc++.h>
  using namespace std;
  template <typename T>
  void read(T &x) {
    x = 0;
    int f = 1;
    char c = getchar();
    for (; !isdigit(c); c = getchar())
      if (c == '-') f = -1;
    for (; isdigit(c); c = getchar()) x = x * 10 + c - '0';
    x = x * f;
  }
  int T, l, r, d[200020];
  int a, b, mid;
  signed main() {
    read(T);
    while (T) {
      T--;
      read(l), read(r);
      if (l == r) {
        cout << 0 << endl;
        continue;
      }
      if (l == 1 and r == 2) {
        cout << 0 << endl;
        continue;
      }
      if (l * 2 >= r) {
        if (r != l * 2)
          cout << r - l << endl;
        else
          cout << l - 1 << endl;
      } else {
        if (r % 2 == 0) r--;
        cout << r % ((r + 1) / 2) << endl;
      }
    }
    return 0;
  }
  