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.
#include <bits/stdc++.h>
using namespace std;
template <typename T>
inline void read(T &x) {
  x = 0;
  T f = 1;
  char c = getchar();
  for (; c < '0' || c > '9'; c = getchar())
    if (c == '-') f = -1;
  for (; c >= '0' && c <= '9'; c = getchar()) x = x * 10 + c - '0';
  x *= f;
}
int main() {
  int t, a, b;
  for (scanf("%d", &t); t--;) {
    scanf("%d %d", &a, &b);
    if (a > (b >> 1))
      printf("%d\n", b % a);
    else
      printf("%d\n", b % ((b >> 1) + 1));
  }
  return 0;
}
