#include <bits/stdc++.h>
using namespace std;
const int N = 400005;
const long long MOD = 1e9 + 7;
const long long M = 1e18;
long long qpow(long long a, long long b, long long m) {
  long long ans = 1;
  while (b) {
    if (b & 1) ans = (ans * a) % m;
    b /= 2;
    a = (a * a) % m;
  }
  return ans;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  int tt = 1;
  cin >> tt;
  for (int t = 1; t <= tt; t++) {
    int l, r;
    cin >> l >> r;
    int k = (r + 1) / 2;
    if (k >= l)
      cout << k - 1;
    else
      cout << r % l;
    cout << endl;
  }
  return 0;
}

### Prompt

Generate 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;
const int N = 400005;
const long long MOD = 1e9 + 7;
const long long M = 1e18;
long long qpow(long long a, long long b, long long m) {
  long long ans = 1;
  while (b) {
    if (b & 1) ans = (ans * a) % m;
    b /= 2;
    a = (a * a) % m;
  }
  return ans;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  int tt = 1;
  cin >> tt;
  for (int t = 1; t <= tt; t++) {
    int l, r;
    cin >> l >> r;
    int k = (r + 1) / 2;
    if (k >= l)
      cout << k - 1;
    else
      cout << r % l;
    cout << endl;
  }
  return 0;
}

```