#include <bits/stdc++.h>
using namespace std;
long long gcd(long long a, long long b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
long long lcm(long long a, long long b) { return a / gcd(a, b) * b; }
const long long INF = (long long)1e5 + 6;
void run() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
}
void solve() {
  int a, b;
  cin >> a >> b;
  if (a == b)
    cout << 0 << endl;
  else {
    if (a > b) {
      if (a / 2 >= b)
        cout << a % ((a / 2) + 1) << endl;
      else
        cout << a % b << endl;
    } else {
      if (b / 2 >= a)
        cout << b % ((b / 2) + 1) << endl;
      else
        cout << b % a << endl;
    }
  }
}
int main() {
  run();
  int n;
  cin >> n;
  while (n--) {
    solve();
  }
  return 0;
}

### Prompt

Your challenge is to write 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;
long long gcd(long long a, long long b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
long long lcm(long long a, long long b) { return a / gcd(a, b) * b; }
const long long INF = (long long)1e5 + 6;
void run() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
}
void solve() {
  int a, b;
  cin >> a >> b;
  if (a == b)
    cout << 0 << endl;
  else {
    if (a > b) {
      if (a / 2 >= b)
        cout << a % ((a / 2) + 1) << endl;
      else
        cout << a % b << endl;
    } else {
      if (b / 2 >= a)
        cout << b % ((b / 2) + 1) << endl;
      else
        cout << b % a << endl;
    }
  }
}
int main() {
  run();
  int n;
  cin >> n;
  while (n--) {
    solve();
  }
  return 0;
}

```