#include <bits/stdc++.h>
using namespace std;
const int mod = 1e9 + 7;
const int M = 5e5 + 5;
const int INF = 2e9;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long int n, m, t, i, j, k, a, b, c;
  cin >> t;
  while (t--) {
    cin >> n >> m;
    if (m % 2 == 0) {
      a = m - 1;
      b = max(m / 2, n);
      c = m % b;
      if (n == m) {
        a = 0;
      }
    } else {
      a = m;
      b = max((m + 1) / 2, n);
      c = 0;
    }
    cout << max(c, a % b) << '\n';
  }
}

### Prompt

In CPP, your task is to solve 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;
const int mod = 1e9 + 7;
const int M = 5e5 + 5;
const int INF = 2e9;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long int n, m, t, i, j, k, a, b, c;
  cin >> t;
  while (t--) {
    cin >> n >> m;
    if (m % 2 == 0) {
      a = m - 1;
      b = max(m / 2, n);
      c = m % b;
      if (n == m) {
        a = 0;
      }
    } else {
      a = m;
      b = max((m + 1) / 2, n);
      c = 0;
    }
    cout << max(c, a % b) << '\n';
  }
}

```