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.
#include <bits/stdc++.h>
using namespace std;
void _print(long long t) { cerr << t; }
void _print(int t) { cerr << t; }
void _print(string t) { cerr << t; }
void _print(char t) { cerr << t; }
void _print(long double t) { cerr << t; }
void _print(double t) { cerr << t; }
void _print(unsigned long long t) { cerr << t; }
template <class T, class V>
void _print(pair<T, V> p);
template <class T>
void _print(vector<T> v);
template <class T>
void _print(set<T> v);
template <class T, class V>
void _print(map<T, V> v);
template <class T>
void _print(multiset<T> v);
template <class T, class V>
void _print(pair<T, V> p) {
  cerr << "{";
  _print(p.first);
  cerr << ",";
  _print(p.second);
  cerr << "}";
}
template <class T>
void _print(vector<T> v) {
  cerr << "[ ";
  for (T i : v) {
    _print(i);
    cerr << " ";
  }
  cerr << "]";
}
template <class T>
void _print(set<T> v) {
  cerr << "[ ";
  for (T i : v) {
    _print(i);
    cerr << " ";
  }
  cerr << "]";
}
template <class T>
void _print(multiset<T> v) {
  cerr << "[ ";
  for (T i : v) {
    _print(i);
    cerr << " ";
  }
  cerr << "]";
}
template <class T, class V>
void _print(map<T, V> v) {
  cerr << "[ ";
  for (auto i : v) {
    _print(i);
    cerr << " ";
  }
  cerr << "]";
}
int main() {
  int t;
  cin >> t;
  while (t--) {
    long long l, r;
    cin >> l >> r;
    if (r < l * 2) {
      cout << r - l << endl;
    } else {
      cout << (r - 1) / 2 << endl;
    }
  }
  return 0;
}
