#include <bits/stdc++.h>
using namespace std;
void PV(vector<long long> v) {
  for (long long i = 0; i < (long long)v.size(); i++) cout << v[i] << " ";
  cout << "\n";
}
void PVV(vector<pair<long long, long long> > v) {
  for (long long i = 0; i < (long long)v.size(); i++)
    cout << v[i].first << " " << v[i].second << "\n";
}
void PA(long long v[], long long n, long long x = 0) {
  for (long long i = x; i < n + x; i++) cout << v[i] << ' ';
  cout << "\n";
}
void IN(long long a[], long long n, long long x = 0) {
  for (long long i = x; i < n + x; i++) cin >> a[i];
}
inline void op() {}
const long long M = 1e9 + 7;
const long long IM = 1e18 + 37;
const long long N = 2e5 + 10;
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  op();
  long long t;
  cin >> t;
  while (t--) {
    long long l, r;
    cin >> l >> r;
    if (l == r) {
      cout << (0) << "\n";
      continue;
    }
    long long a = (r + 1) / 2;
    long long ans = (a + a - 1) % a;
    if (a < l) {
      ans = (r % l);
    }
    cout << (ans) << "\n";
  }
  return 0;
}

### Prompt

In cpp, your task is to solve 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;
void PV(vector<long long> v) {
  for (long long i = 0; i < (long long)v.size(); i++) cout << v[i] << " ";
  cout << "\n";
}
void PVV(vector<pair<long long, long long> > v) {
  for (long long i = 0; i < (long long)v.size(); i++)
    cout << v[i].first << " " << v[i].second << "\n";
}
void PA(long long v[], long long n, long long x = 0) {
  for (long long i = x; i < n + x; i++) cout << v[i] << ' ';
  cout << "\n";
}
void IN(long long a[], long long n, long long x = 0) {
  for (long long i = x; i < n + x; i++) cin >> a[i];
}
inline void op() {}
const long long M = 1e9 + 7;
const long long IM = 1e18 + 37;
const long long N = 2e5 + 10;
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  op();
  long long t;
  cin >> t;
  while (t--) {
    long long l, r;
    cin >> l >> r;
    if (l == r) {
      cout << (0) << "\n";
      continue;
    }
    long long a = (r + 1) / 2;
    long long ans = (a + a - 1) % a;
    if (a < l) {
      ans = (r % l);
    }
    cout << (ans) << "\n";
  }
  return 0;
}

```