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;
  int ured() {
    int re = 0, ch;
    do {
      ch = getchar();
    } while ('9' < ch || ch < '0');
    do {
      re = re * 10 + (ch ^ '0');
    } while ('0' <= (ch = getchar()) && ch <= '9');
    return re;
  }
  void uwit(int da) {
    int cn = 0, ch[21];
    do {
      ch[++cn] = da - da / 10 * 10;
    } while (da /= 10);
    do {
      putchar('0' ^ ch[cn]);
    } while (--cn);
  }
  inline void dmax(int& fr, int to) {
    if (fr < to) {
      fr = to;
    }
  }
  inline void dmin(int& fr, int to) {
    if (fr > to) {
      fr = to;
    }
  }
  inline int mabs(int da) { return da >= 0 ? da : -da; }
  const int _maxn = 200011, _mods = 1000000007;
  int t, n, m, pdif;
  char s[_maxn];
  int main() {
    t = ured();
    while (t--) {
      n = ured();
      m = ured();
      if (m >= 2 * n) {
        uwit((m + 1) / 2 - 1);
      } else {
        uwit(m - n);
      }
      putchar('\n');
    }
    return 0;
  }
  