#include <bits/stdc++.h>
using namespace std;
struct _ {
  ios_base::Init i;
  _() {
    cin.sync_with_stdio(0);
    cin.tie(0);
    cout << fixed << setprecision(10);
  }
} ___;
template <typename Arg1>
void __f(const char* name, Arg1&& arg1) {
  cerr << name << " : " << arg1 << endl;
}
template <typename Arg1, typename... Args>
void __f(const char* names, Arg1&& arg1, Args&&... args) {
  const char* comma = strchr(names + 1, ',');
  cerr.write(names, comma - names) << " : " << arg1 << "  ";
  __f(comma + 1, args...);
}
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
long long int my_rand(long long int l, long long int r) {
  return uniform_int_distribution<long long int>(l, r)(rng);
}
signed main() {
  long long int __, _ = 0;
  cin >> __;
  for (; _++ < __;) {
    long long int l, r;
    cin >> l >> r;
    long long int lo = l, hi = r;
    while (lo < hi) {
      long long int mid = (lo + hi + 1) >> 1;
      if (mid * 2 - 1 >= l and mid * 2 - 1 <= r)
        lo = mid;
      else
        hi = mid - 1;
    }
    if (lo * 2 - 1 >= l and lo * 2 - 1 <= r) {
      cout << lo - 1 << '\n';
    } else
      cout << r % l << '\n';
  }
}

### Prompt

Develop a solution in Cpp 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;
struct _ {
  ios_base::Init i;
  _() {
    cin.sync_with_stdio(0);
    cin.tie(0);
    cout << fixed << setprecision(10);
  }
} ___;
template <typename Arg1>
void __f(const char* name, Arg1&& arg1) {
  cerr << name << " : " << arg1 << endl;
}
template <typename Arg1, typename... Args>
void __f(const char* names, Arg1&& arg1, Args&&... args) {
  const char* comma = strchr(names + 1, ',');
  cerr.write(names, comma - names) << " : " << arg1 << "  ";
  __f(comma + 1, args...);
}
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
long long int my_rand(long long int l, long long int r) {
  return uniform_int_distribution<long long int>(l, r)(rng);
}
signed main() {
  long long int __, _ = 0;
  cin >> __;
  for (; _++ < __;) {
    long long int l, r;
    cin >> l >> r;
    long long int lo = l, hi = r;
    while (lo < hi) {
      long long int mid = (lo + hi + 1) >> 1;
      if (mid * 2 - 1 >= l and mid * 2 - 1 <= r)
        lo = mid;
      else
        hi = mid - 1;
    }
    if (lo * 2 - 1 >= l and lo * 2 - 1 <= r) {
      cout << lo - 1 << '\n';
    } else
      cout << r % l << '\n';
  }
}

```