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.
#include <bits/stdc++.h>
using namespace std;
long long int l, r;
long long int fun(long long int x) {
  long long int ans = 0;
  if (x < l) {
    return 0;
  }
  if (2 * x - 1 <= r) {
    return x - 1;
  }
  return r - x;
}
void solve() {
  cin >> l >> r;
  long long int ans = 0;
  long long int x = r / 2;
  x++;
  ans = max(ans, fun(x));
  x--;
  ans = max(ans, fun(x));
  x = l;
  ans = max(ans, fun(x));
  cout << ans << "\n";
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long int t, q;
  q = 1;
  cin >> q;
  for (t = 1; t <= q; t++) {
    solve();
  }
}
