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;
  const int INF = 0x3f3f3f3f;
  const long long LINF = 0x3f3f3f3f3f3f3f3f;
  const double eps = 1e-12;
  const double PI = acos(-1.0);
  const double angcst = PI / 180.0;
  const long long mod = 998244353;
  long long max_3(long long a, long long b, long long c) {
    if (a > b && a > c) return a;
    if (b > c) return b;
    return c;
  }
  long long min_3(long long a, long long b, long long c) {
    if (a < b && a < c) return a;
    if (b < c) return b;
    return c;
  }
  long long gcd(long long a, long long b) { return b == 0 ? a : gcd(b, a % b); }
  long long lcm(long long a, long long b) { return a / gcd(a, b) * b; }
  long long qpow(long long a, long long n) {
    long long r = 1;
    while (n) {
      if (n & 1) r = (r * a) % mod;
      n >>= 1;
      a = (a * a) % mod;
    }
    return r;
  }
  long long qmul(long long a, long long b) {
    long long s = (long double)a / mod * b;
    s = a * b - s * mod;
    if (s < 0) s += mod;
    if (s >= mod) s -= mod;
    return s;
  }
  template <typename _Tp>
  inline _Tp read(_Tp& x) {
    char c11 = getchar(), ob = 0;
    x = 0;
    while (c11 ^ '-' && !isdigit(c11)) c11 = getchar();
    if (c11 == '-') c11 = getchar(), ob = 1;
    while (isdigit(c11)) x = x * 10 + c11 - '0', c11 = getchar();
    if (ob) x = -x;
    return x;
  }
  const int maxn = 10010;
  long long t, a, b;
  int main() {
    cin >> t;
    while (t--) {
      cin >> a >> b;
      if (b % 2 == 0 && a == b / 2) {
        cout << a - 1 << endl;
        continue;
      }
      if (b % 2 == 1 && a == b / 2) {
        cout << a << endl;
        continue;
      }
      if (b % 2 == 1) {
        int x = b / 2, y = b / 2 + 1;
        if (x >= a) {
          cout << x << endl;
        } else {
          cout << b - a << endl;
        }
      } else {
        int x = b / 2 - 1, y = b / 2 + 1;
        if (x >= a) {
          cout << x << endl;
        } else {
          cout << b - a << endl;
        }
      }
    }
    return 0;
  }
  