#include <bits/stdc++.h>
using namespace std;
template <typename A>
ostream &operator<<(ostream &cout, vector<A> const &v);
template <typename A, typename B>
ostream &operator<<(ostream &cout, pair<A, B> const &p) {
  return cout << '(' << p.f << ',' << ' ' << p.s << ')';
}
template <typename A>
ostream &operator<<(ostream &cout, vector<A> const &v) {
  cout << '[';
  for (int i = 0; i < v.size(); i++) {
    if (i) cout << ',';
    cout << v[i];
  }
  return cout << ']' << endl;
}
const long long mod = 1000000007;
void solve() {
  long long a, b;
  cin >> a >> b;
  long long n = b / 2 + 1;
  cout << b % max(a, n) << "\n";
}
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int tc = 1;
  cin >> tc;
  while (tc--) {
    solve();
  }
  return 0;
}

### Prompt

Generate a Cpp solution to the following problem:

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.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
template <typename A>
ostream &operator<<(ostream &cout, vector<A> const &v);
template <typename A, typename B>
ostream &operator<<(ostream &cout, pair<A, B> const &p) {
  return cout << '(' << p.f << ',' << ' ' << p.s << ')';
}
template <typename A>
ostream &operator<<(ostream &cout, vector<A> const &v) {
  cout << '[';
  for (int i = 0; i < v.size(); i++) {
    if (i) cout << ',';
    cout << v[i];
  }
  return cout << ']' << endl;
}
const long long mod = 1000000007;
void solve() {
  long long a, b;
  cin >> a >> b;
  long long n = b / 2 + 1;
  cout << b % max(a, n) << "\n";
}
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int tc = 1;
  cin >> tc;
  while (tc--) {
    solve();
  }
  return 0;
}

```