#include <bits/stdc++.h>
using namespace std;
void Pagla() {
  int i, j;
  long long int a, b;
  cin >> a >> b;
  if (a == b) {
    cout << 0 << "\n";
    return;
  }
  if (b - a == 1) {
    if (b % a == 0)
      cout << 0 << "\n";
    else
      cout << 1 << "\n";
    return;
  }
  if (b / 2 >= a) {
    long long int ak = ceil(b * 1.00 / 2);
    long long int mx = 0;
    mx = max(mx, b % ak);
    while (1) {
      if (b % ak != 0) {
        break;
      }
      mx = max(mx, b % ak);
      ak++;
      if (ak >= b) {
        cout << mx << "\n";
        return;
      }
    }
    cout << max(mx, b % ak) << "\n";
  } else {
    if (b % a != 0) {
      cout << b % a << "\n";
    } else {
      if (a != 1)
        cout << b % (a - 1) << "\n";
      else
        cout << b % a << "\n";
    }
  }
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int t;
  t = 1;
  cin >> t;
  while (t--) {
    Pagla();
  }
  return 0;
}

### Prompt

Please formulate 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;
void Pagla() {
  int i, j;
  long long int a, b;
  cin >> a >> b;
  if (a == b) {
    cout << 0 << "\n";
    return;
  }
  if (b - a == 1) {
    if (b % a == 0)
      cout << 0 << "\n";
    else
      cout << 1 << "\n";
    return;
  }
  if (b / 2 >= a) {
    long long int ak = ceil(b * 1.00 / 2);
    long long int mx = 0;
    mx = max(mx, b % ak);
    while (1) {
      if (b % ak != 0) {
        break;
      }
      mx = max(mx, b % ak);
      ak++;
      if (ak >= b) {
        cout << mx << "\n";
        return;
      }
    }
    cout << max(mx, b % ak) << "\n";
  } else {
    if (b % a != 0) {
      cout << b % a << "\n";
    } else {
      if (a != 1)
        cout << b % (a - 1) << "\n";
      else
        cout << b % a << "\n";
    }
  }
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int t;
  t = 1;
  cin >> t;
  while (t--) {
    Pagla();
  }
  return 0;
}

```