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 minimum(int x, int y) { return (x < y ? x : y); }
  template <typename type>
  struct even {
    bool operator()(const type &x, const type &y) const { return x % 2 < y % 2; }
  };
  template <typename type>
  struct odd {
    bool operator()(const type &x, const type &y) const { return x % 2 > y % 2; }
  };
  template <typename type>
  struct abslouteValue {
    bool operator()(const type &x, const type &y) const {
      return abs(x) > abs(y);
    }
  };
  bool flusher() {
    string s;
    fflush(stdout);
    getline(cin, s);
    return s != "";
  }
  void insertionSortAlgorithm(int arr[], int n) {
    int key, j;
    for (int i = 1; i < n; i++) {
      key = arr[i];
      j = i - 1;
      while (j >= 0 && arr[j] > key) {
        arr[j + 1] = arr[j];
        j = j - 1;
      }
      arr[j + 1] = key;
    }
  }
  template <class T>
  T sum(T x) {
    string s = to_string(x);
    T sumOfString = 0;
    int n = s.size();
    for (long long int i = 0; i < n; i++) {
      sumOfString += (T)s[i] - '0';
    }
    return sumOfString;
  }
  bool compare(const pair<double, double> a, const pair<double, double> b) {
    if (a.first == b.first) {
      return b.second > a.second;
    } else {
      return a.first > b.first;
    }
  }
  bool abslout(const int &x, const int &y) { return abs(x) > abs(y); }
  bool isPrime(long long int x) {
    if (x == 1) return false;
    bool yes = true;
    if (x == 2) {
      return yes;
    } else {
      for (long long int i = 2; i < x; i++) {
        if (x % i == 0) {
          yes = false;
          break;
        }
      }
    }
    return yes;
  }
  void build(int l, int r, vector<int> const &a, vector<int> &res, int d = 0) {
    if (r < l) {
      return;
    }
    if (l == r) {
      res[l] = d;
      return;
    }
    int temp = l;
    for (int i = l + 1; i <= r; i++) {
      if (a[temp] < a[i]) {
        temp = i;
      }
    }
    res[temp] = d;
    build(l, temp - 1, a, res, d + 1);
    build(temp + 1, r, a, res, d + 1);
  }
  bool binarySearch(int l, int r, long long int *a, long long int k) {
    while (l <= r) {
      long long int mid = (l + r) / 2;
      if (a[mid] == k) {
        return true;
      } else if (a[mid] > k) {
        r = mid - 1;
      } else {
        l = mid + 1;
      }
    }
    return false;
  }
  int gcd(int a, int b) {
    if (b == 0) {
      return a;
    }
    return gcd(b, a % b);
  }
  vector<long long int> luckyPackage;
  void bruteForce(long long int n, long long int four, long long int seven) {
    if (n > 1e10) return;
    if (four == seven) {
      luckyPackage.push_back(n);
    }
    bruteForce(n * 10 + 4, four + 1, seven);
    bruteForce(n * 10 + 7, four, seven + 1);
  }
  long long int gcd(long long int a, long long int b) {
    if (b == 0) return a;
    return gcd(b, a % b);
  }
  int main() {
    ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0);
    int t;
    cin >> t;
    while (t--) {
      int a, b;
      cin >> a >> b;
      cout << b % (max(b / 2 + 1, a)) << endl;
    }
    return 0;
  }
  