#include <bits/stdc++.h>
using namespace std;
const long long MAX = 5005, INF = 1e15;
long long MOD = 998244353;
template <typename T>
void c_min(T &a, const T b) {
  if (b <= a) a = b;
}
template <typename T>
void c_max(T &a, const T b) {
  if (a <= b) a = b;
}
template <typename T>
T mexp(T a, long long e) {
  T res = 1;
  while (e) {
    if (e & 1) res *= a;
    a *= a;
    e >>= 1;
  }
  return res;
}
void test_case(long long tc = 0) {
  int l, r;
  cin >> l >> r;
  int x = r / 2 + 1;
  int ans = r % l;
  if (x >= l) {
    ans = max(ans, r % x);
  }
  cout << ans << '\n';
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long T = 1, tc = 0;
  cin >> T;
  for (; tc < T; tc++) {
    test_case(tc);
  }
  return 0;
}

### Prompt

Please formulate 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;
const long long MAX = 5005, INF = 1e15;
long long MOD = 998244353;
template <typename T>
void c_min(T &a, const T b) {
  if (b <= a) a = b;
}
template <typename T>
void c_max(T &a, const T b) {
  if (a <= b) a = b;
}
template <typename T>
T mexp(T a, long long e) {
  T res = 1;
  while (e) {
    if (e & 1) res *= a;
    a *= a;
    e >>= 1;
  }
  return res;
}
void test_case(long long tc = 0) {
  int l, r;
  cin >> l >> r;
  int x = r / 2 + 1;
  int ans = r % l;
  if (x >= l) {
    ans = max(ans, r % x);
  }
  cout << ans << '\n';
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long T = 1, tc = 0;
  cin >> T;
  for (; tc < T; tc++) {
    test_case(tc);
  }
  return 0;
}

```