#include <bits/stdc++.h>
#pragma GCC optimize("Ofast")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,avx2,fma")
#pragma GCC optimize("unroll-loops")
using namespace std;
vector<string> tokenizer(string str, char ch) {
  std::istringstream var((str));
  vector<string> v;
  string t;
  while (getline((var), t, (ch))) {
    v.push_back(t);
  }
  return v;
}
void err(istream_iterator<string> it) {}
template <typename T, typename... Args>
void err(istream_iterator<string> it, T a, Args... args) {
  cout << *it << " = " << a << "\n";
  err(++it, args...);
}
void file_i_o() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
}
int main(int argc, char const *argv[]) {
  file_i_o();
  long long int t;
  cin >> t;
  while (t--) {
    long long int l, r;
    cin >> l >> r;
    if (r >= 2 * l - 1) {
      if (r % 2 == 0) r--;
      cout << r % ((r + 1) / 2) << "\n";
    } else {
      cout << r % l << "\n";
    }
  }
  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>
#pragma GCC optimize("Ofast")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,avx2,fma")
#pragma GCC optimize("unroll-loops")
using namespace std;
vector<string> tokenizer(string str, char ch) {
  std::istringstream var((str));
  vector<string> v;
  string t;
  while (getline((var), t, (ch))) {
    v.push_back(t);
  }
  return v;
}
void err(istream_iterator<string> it) {}
template <typename T, typename... Args>
void err(istream_iterator<string> it, T a, Args... args) {
  cout << *it << " = " << a << "\n";
  err(++it, args...);
}
void file_i_o() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
}
int main(int argc, char const *argv[]) {
  file_i_o();
  long long int t;
  cin >> t;
  while (t--) {
    long long int l, r;
    cin >> l >> r;
    if (r >= 2 * l - 1) {
      if (r % 2 == 0) r--;
      cout << r % ((r + 1) / 2) << "\n";
    } else {
      cout << r % l << "\n";
    }
  }
  return 0;
}

```