#include <bits/stdc++.h>
using namespace std;
const long long maxn = 4e5 + 10;
const long long mod = 998244353;
long long k, A, h;
long long qpow(long long x, long long n) {
  long long tmp = x % mod;
  long long a = 1;
  while (n) {
    if (n % 2) a = (a * tmp) % mod;
    n = n / 2;
    tmp = (tmp * tmp) % mod;
  }
  return a;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int tc = 1;
  cin >> tc;
  while (tc--) {
    long long a, b;
    cin >> a >> b;
    if (a <= (b - 1) / 2)
      cout << (b - 1) / 2 << "\n";
    else
      cout << min(b - a, (b - 1) / 2) << "\n";
  }
}

### Prompt

Please provide a cpp coded solution to the problem described below:

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 maxn = 4e5 + 10;
const long long mod = 998244353;
long long k, A, h;
long long qpow(long long x, long long n) {
  long long tmp = x % mod;
  long long a = 1;
  while (n) {
    if (n % 2) a = (a * tmp) % mod;
    n = n / 2;
    tmp = (tmp * tmp) % mod;
  }
  return a;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int tc = 1;
  cin >> tc;
  while (tc--) {
    long long a, b;
    cin >> a >> b;
    if (a <= (b - 1) / 2)
      cout << (b - 1) / 2 << "\n";
    else
      cout << min(b - a, (b - 1) / 2) << "\n";
  }
}

```