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, typename U>
  inline void amin(T &x, U y) {
    if (y < x) x = y;
  }
  template <typename T, typename U>
  inline void amax(T &x, U y) {
    if (x < y) x = y;
  }
  template <typename T>
  inline void write(T x) {
    long long i = 20;
    char buf[21];
    buf[20] = '\n';
    do {
      buf[--i] = x % 10 + '0';
      x /= 10;
    } while (x);
    do {
      putchar(buf[i]);
    } while (buf[i++] != '\n');
  }
  template <typename T>
  inline T readInt() {
    T n = 0, s = 1;
    char p = getchar();
    if (p == '-') s = -1;
    while ((p < '0' || p > '9') && p != EOF && p != '-') p = getchar();
    if (p == '-') s = -1, p = getchar();
    while (p >= '0' && p <= '9') {
      n = (n << 3) + (n << 1) + (p - '0');
      p = getchar();
    }
    return n * s;
  }
  long long gcd(long long a, long long b) {
    if (a == 0) return b;
    if (b == 0) return a;
    if (a == b) return a;
    if (a > b)
      return gcd(a % b, b);
    else
      return gcd(a, b % a);
  }
  long long exp(long long x, long long n) {
    if (n == 0) return 1;
    if (n == 1) return x % 1000000007;
    if (n % 2 == 0)
      return exp((x * x) % 1000000007, n / 2);
    else
      return exp(x * exp((x * x) % 1000000007, n / 2));
  }
  bool sortbysec(const pair<long long, long long> &a,
                 const pair<long long, long long> &b) {
    return (a.second < b.second);
  }
  void solve() {
    long long l, r;
    cin >> l >> r;
    if (2 * l <= r) {
      if (r % 2 == 0)
        cout << r / 2 - 1 << "\n";
      else
        cout << r / 2 << "\n";
    } else {
      cout << r - l << "\n";
    }
  }
  int32_t main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);
    long long t = 1;
    cin >> t;
    while (t--) {
      solve();
    }
  }
  