#include <bits/stdc++.h>
using namespace std;
void edit(vector<int> &skill, int n) {
  for (int i = 0; i < n; i++) {
    int x;
    cin >> x;
    skill.push_back(x);
  }
}
void editl(vector<long long> &skill, long long n) {
  for (long long i = 0; i < n; i++) {
    long long x;
    cin >> x;
    skill.push_back(x);
  }
}
long long gcd(long long a, long long b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
long long binpow(long long a, long long b) {
  long long res = 1;
  while (b > 0) {
    if (b & 1) res = res * a;
    a = a * a;
    b >>= 1;
  }
  return res;
}
bool isGoogles = 0;
bool testcase = 1;
void achilles() {
  long long a, b;
  cin >> a >> b;
  if (b % 2) {
    if (a <= b / 2 + 1) {
      cout << b / 2 << "\n";
      return;
    } else {
      cout << b % a << "\n";
      return;
    }
  } else {
    if (a <= b / 2 + 1) {
      cout << b / 2 - 1 << "\n";
      return;
    } else {
      cout << b % a << "\n";
      return;
    }
  }
}
int main() {
  ios ::sync_with_stdio(0);
  cin.tie(0);
  long long t = 1, w = 1;
  if (testcase) cin >> t;
  while (t--) {
    if (isGoogles) {
      cout << "Case #" << w << ": ";
    }
    achilles();
    w++;
  }
  return 0;
}

### Prompt

Generate 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;
void edit(vector<int> &skill, int n) {
  for (int i = 0; i < n; i++) {
    int x;
    cin >> x;
    skill.push_back(x);
  }
}
void editl(vector<long long> &skill, long long n) {
  for (long long i = 0; i < n; i++) {
    long long x;
    cin >> x;
    skill.push_back(x);
  }
}
long long gcd(long long a, long long b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
long long binpow(long long a, long long b) {
  long long res = 1;
  while (b > 0) {
    if (b & 1) res = res * a;
    a = a * a;
    b >>= 1;
  }
  return res;
}
bool isGoogles = 0;
bool testcase = 1;
void achilles() {
  long long a, b;
  cin >> a >> b;
  if (b % 2) {
    if (a <= b / 2 + 1) {
      cout << b / 2 << "\n";
      return;
    } else {
      cout << b % a << "\n";
      return;
    }
  } else {
    if (a <= b / 2 + 1) {
      cout << b / 2 - 1 << "\n";
      return;
    } else {
      cout << b % a << "\n";
      return;
    }
  }
}
int main() {
  ios ::sync_with_stdio(0);
  cin.tie(0);
  long long t = 1, w = 1;
  if (testcase) cin >> t;
  while (t--) {
    if (isGoogles) {
      cout << "Case #" << w << ": ";
    }
    achilles();
    w++;
  }
  return 0;
}

```