#include <bits/stdc++.h>
using namespace std;
const long long inf = 1e15;
long long l, r;
void Input() { cin >> l >> r; }
void Solve() {
  if (r <= 10) {
    long long Res = 0;
    for (long long a = l; a <= r; ++a) {
      for (long long b = l; b <= a; ++b) Res = max(Res, a % b);
    }
    cout << Res;
  } else {
    long long m = max(r / 2 + 1, l);
    cout << max(r % l, r % m);
  }
  cout << '\n';
}
int main() {
  if (fopen("trash.inp", "r"))
    freopen("trash.inp", "r", stdin), freopen("trash.out", "w", stdout);
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int Test;
  cin >> Test;
  while (Test--) {
    Input();
    Solve();
  }
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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 long long inf = 1e15;
long long l, r;
void Input() { cin >> l >> r; }
void Solve() {
  if (r <= 10) {
    long long Res = 0;
    for (long long a = l; a <= r; ++a) {
      for (long long b = l; b <= a; ++b) Res = max(Res, a % b);
    }
    cout << Res;
  } else {
    long long m = max(r / 2 + 1, l);
    cout << max(r % l, r % m);
  }
  cout << '\n';
}
int main() {
  if (fopen("trash.inp", "r"))
    freopen("trash.inp", "r", stdin), freopen("trash.out", "w", stdout);
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int Test;
  cin >> Test;
  while (Test--) {
    Input();
    Solve();
  }
}

```