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.
def solve(Test_cases):
    Resultant_list = []
    for i in range(Test_cases):
        # Start the code
        string = input()
        list_me = list(string.split())
        map_object = map(int, list_me)
        list_me = list(map_object)
        l, r = list_me[0], list_me[1]

        if r == l:
            Resultant_list.append(0)
        elif r % 2 != 0:
            # r is odd
            value_1 = (r + 1) // 2
            if value_1 >= l:
                Resultant_list.append(r % value_1)
            else:
                Resultant_list.append(r % l)
        else:
            # r is even
            value_2 = (r // 2) + 1
            if value_2 >= l:
                Resultant_list.append(r % value_2)
            else:
                Resultant_list.append(r % l)
    return Resultant_list


Test_cases = int(input())
if Test_cases < 1 or Test_cases > pow(10, 4):
    exit()
Result = solve(Test_cases)
for i in Result:
    print(i)
