#include <bits/stdc++.h>
using namespace std;
long long prime(int n) {
  if (n == 1) return 0;
  if (n == 2 || n == 3) return 1;
  if (n % 2 == 0 || n % 3 == 0) return 0;
  for (long long i = 5; i * i <= n; i += 6) {
    if (n % i == 0 || n % (i + 2) == 0) {
      return 0;
    }
  }
  return 1;
}
void solve1() {
  long long a, b;
  cin >> a >> b;
  if (a == b) {
    cout << 0 << endl;
  } else {
    long long val = 0;
    val = b / 2 + 1;
    val = max(val, a);
    cout << b % val << endl;
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long t;
  cin >> t;
  for (long long i = 0; i < t; i++) {
    solve1();
  }
  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 prime(int n) {
  if (n == 1) return 0;
  if (n == 2 || n == 3) return 1;
  if (n % 2 == 0 || n % 3 == 0) return 0;
  for (long long i = 5; i * i <= n; i += 6) {
    if (n % i == 0 || n % (i + 2) == 0) {
      return 0;
    }
  }
  return 1;
}
void solve1() {
  long long a, b;
  cin >> a >> b;
  if (a == b) {
    cout << 0 << endl;
  } else {
    long long val = 0;
    val = b / 2 + 1;
    val = max(val, a);
    cout << b % val << endl;
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long t;
  cin >> t;
  for (long long i = 0; i < t; i++) {
    solve1();
  }
  return 0;
}

```