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>
  inline void read(T &x) {
    x = 0;
    T f = 1;
    char c = getchar();
    for (; c < '0' || c > '9'; c = getchar())
      if (c == '-') f = -1;
    for (; c >= '0' && c <= '9'; c = getchar()) x = x * 10 + c - '0';
    x *= f;
  }
  int main() {
    int t, a, b;
    for (scanf("%d", &t); t--;) {
      scanf("%d %d", &a, &b);
      if (a > (b >> 1))
        printf("%d\n", b % a);
      else
        printf("%d\n", b % ((b >> 1) + 1));
    }
    return 0;
  }
  