#include <bits/stdc++.h>
using namespace std;
int main() {
  long long t, i, l, r;
  cin >> t;
  for (int i = 0; i < t; i++) {
    cin >> l >> r;
    long long x = r / 2;
    if (l > x)
      cout << r % l << endl;
    else
      cout << r % (x + 1) << endl;
  }
  return 0;
}

### Prompt

Your task is to create 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;
int main() {
  long long t, i, l, r;
  cin >> t;
  for (int i = 0; i < t; i++) {
    cin >> l >> r;
    long long x = r / 2;
    if (l > x)
      cout << r % l << endl;
    else
      cout << r % (x + 1) << endl;
  }
  return 0;
}

```