#include <bits/stdc++.h>
using namespace std;
template <typename T>
istream &operator>>(istream &in, vector<T> &v) {
  for (auto &i : v) cin >> i;
  return in;
}
template <typename T>
ostream &operator<<(ostream &out, vector<T> &v) {
  for (auto &i : v) cout << i << " ";
  return out;
}
void solve() {
  long long int i, j, n, a, b, c, d, m, k, sum = 0, ans = 0, cnt = 0, r = 0,
                                           e = 0, x = 0, y = 0;
  string str, s1, om, s;
  map<long long int, long long int> mp, mp1;
  set<long long int> st;
  cin >> a >> b;
  c = b - a;
  b--;
  if ((b + 1) >= 2 * a)
    ans = b / 2;
  else
    ans = c;
  cout << ans << "\n";
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  long long int tc;
  cin >> tc;
  while (tc--) {
    solve();
  }
  return 0;
}

### Prompt

Your task is to create 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;
template <typename T>
istream &operator>>(istream &in, vector<T> &v) {
  for (auto &i : v) cin >> i;
  return in;
}
template <typename T>
ostream &operator<<(ostream &out, vector<T> &v) {
  for (auto &i : v) cout << i << " ";
  return out;
}
void solve() {
  long long int i, j, n, a, b, c, d, m, k, sum = 0, ans = 0, cnt = 0, r = 0,
                                           e = 0, x = 0, y = 0;
  string str, s1, om, s;
  map<long long int, long long int> mp, mp1;
  set<long long int> st;
  cin >> a >> b;
  c = b - a;
  b--;
  if ((b + 1) >= 2 * a)
    ans = b / 2;
  else
    ans = c;
  cout << ans << "\n";
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  long long int tc;
  cin >> tc;
  while (tc--) {
    solve();
  }
  return 0;
}

```