#include <bits/stdc++.h>
using namespace std;
long long max2(long long x, long long y) {
  if (x > y)
    return x;
  else
    return y;
}
long long min2(long long x, long long y) {
  if (x < y)
    return x;
  else
    return y;
}
unsigned long long min3(unsigned long long x, unsigned long long y) {
  if (x < y)
    return x;
  else
    return y;
}
long double min(long double x, long double y) {
  if (x < y)
    return x;
  else
    return y;
}
long long gcd(long long a, long long b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
long long parent(long long i) { return (i - 1) / 2; }
long long left(long long i) { return i * 2 + 1; }
long long right(long long i) { return 2 * i + 2; }
bool within(long long x, long long l, long long r) {
  return ((x >= l) && (x <= r)) || ((x <= l) && (x >= r));
}
void solve(long long t) {
  unsigned long long l, r;
  cin >> l >> r;
  long long ans = r / 2;
  if (ans >= l) {
    while (2 * (ans + 1) - 1 <= r) ans++;
    cout << ans - 1 << endl;
  } else
    cout << r % l << endl;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long t = 1;
  int q = 0;
  cin >> t;
  while (t--) {
    q++;
    solve(q);
  }
  return 0;
}

### 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;
long long max2(long long x, long long y) {
  if (x > y)
    return x;
  else
    return y;
}
long long min2(long long x, long long y) {
  if (x < y)
    return x;
  else
    return y;
}
unsigned long long min3(unsigned long long x, unsigned long long y) {
  if (x < y)
    return x;
  else
    return y;
}
long double min(long double x, long double y) {
  if (x < y)
    return x;
  else
    return y;
}
long long gcd(long long a, long long b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
long long parent(long long i) { return (i - 1) / 2; }
long long left(long long i) { return i * 2 + 1; }
long long right(long long i) { return 2 * i + 2; }
bool within(long long x, long long l, long long r) {
  return ((x >= l) && (x <= r)) || ((x <= l) && (x >= r));
}
void solve(long long t) {
  unsigned long long l, r;
  cin >> l >> r;
  long long ans = r / 2;
  if (ans >= l) {
    while (2 * (ans + 1) - 1 <= r) ans++;
    cout << ans - 1 << endl;
  } else
    cout << r % l << endl;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long t = 1;
  int q = 0;
  cin >> t;
  while (t--) {
    q++;
    solve(q);
  }
  return 0;
}

```