#include <bits/stdc++.h>
using namespace std;
template <typename T, typename U>
inline void amin(T &x, U y) {
  if (y < x) x = y;
}
template <typename T, typename U>
inline void amax(T &x, U y) {
  if (x < y) x = y;
}
template <typename T>
inline void write(T x) {
  long long i = 20;
  char buf[21];
  buf[20] = '\n';
  do {
    buf[--i] = x % 10 + '0';
    x /= 10;
  } while (x);
  do {
    putchar(buf[i]);
  } while (buf[i++] != '\n');
}
template <typename T>
inline T readInt() {
  T n = 0, s = 1;
  char p = getchar();
  if (p == '-') s = -1;
  while ((p < '0' || p > '9') && p != EOF && p != '-') p = getchar();
  if (p == '-') s = -1, p = getchar();
  while (p >= '0' && p <= '9') {
    n = (n << 3) + (n << 1) + (p - '0');
    p = getchar();
  }
  return n * s;
}
long long gcd(long long a, long long b) {
  if (a == 0) return b;
  if (b == 0) return a;
  if (a == b) return a;
  if (a > b)
    return gcd(a % b, b);
  else
    return gcd(a, b % a);
}
long long exp(long long x, long long n) {
  if (n == 0) return 1;
  if (n == 1) return x % 1000000007;
  if (n % 2 == 0)
    return exp((x * x) % 1000000007, n / 2);
  else
    return exp(x * exp((x * x) % 1000000007, n / 2));
}
bool sortbysec(const pair<long long, long long> &a,
               const pair<long long, long long> &b) {
  return (a.second < b.second);
}
void solve() {
  long long l, r;
  cin >> l >> r;
  if (2 * l <= r) {
    if (r % 2 == 0)
      cout << r / 2 - 1 << "\n";
    else
      cout << r / 2 << "\n";
  } else {
    cout << r - l << "\n";
  }
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long t = 1;
  cin >> t;
  while (t--) {
    solve();
  }
}

### Prompt

Your challenge is to write 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;
template <typename T, typename U>
inline void amin(T &x, U y) {
  if (y < x) x = y;
}
template <typename T, typename U>
inline void amax(T &x, U y) {
  if (x < y) x = y;
}
template <typename T>
inline void write(T x) {
  long long i = 20;
  char buf[21];
  buf[20] = '\n';
  do {
    buf[--i] = x % 10 + '0';
    x /= 10;
  } while (x);
  do {
    putchar(buf[i]);
  } while (buf[i++] != '\n');
}
template <typename T>
inline T readInt() {
  T n = 0, s = 1;
  char p = getchar();
  if (p == '-') s = -1;
  while ((p < '0' || p > '9') && p != EOF && p != '-') p = getchar();
  if (p == '-') s = -1, p = getchar();
  while (p >= '0' && p <= '9') {
    n = (n << 3) + (n << 1) + (p - '0');
    p = getchar();
  }
  return n * s;
}
long long gcd(long long a, long long b) {
  if (a == 0) return b;
  if (b == 0) return a;
  if (a == b) return a;
  if (a > b)
    return gcd(a % b, b);
  else
    return gcd(a, b % a);
}
long long exp(long long x, long long n) {
  if (n == 0) return 1;
  if (n == 1) return x % 1000000007;
  if (n % 2 == 0)
    return exp((x * x) % 1000000007, n / 2);
  else
    return exp(x * exp((x * x) % 1000000007, n / 2));
}
bool sortbysec(const pair<long long, long long> &a,
               const pair<long long, long long> &b) {
  return (a.second < b.second);
}
void solve() {
  long long l, r;
  cin >> l >> r;
  if (2 * l <= r) {
    if (r % 2 == 0)
      cout << r / 2 - 1 << "\n";
    else
      cout << r / 2 << "\n";
  } else {
    cout << r - l << "\n";
  }
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long t = 1;
  cin >> t;
  while (t--) {
    solve();
  }
}

```