You are given two integers a and b, and q queries. The i-th query consists of two numbers l_i and r_i, and the answer to it is the number of integers x such that l_i ≤ x ≤ r_i, and ((x mod a) mod b) ≠ ((x mod b) mod a). Calculate the answer for each query.

Recall that y mod z is the remainder of the division of y by z. For example, 5 mod 3 = 2, 7 mod 8 = 7, 9 mod 4 = 1, 9 mod 9 = 0.

Input

The first line contains one integer t (1 ≤ t ≤ 100) — the number of test cases. Then the test cases follow.

The first line of each test case contains three integers a, b and q (1 ≤ a, b ≤ 200; 1 ≤ q ≤ 500).

Then q lines follow, each containing two integers l_i and r_i (1 ≤ l_i ≤ r_i ≤ 10^{18}) for the corresponding query.

Output

For each test case, print q integers — the answers to the queries of this test case in the order they appear.

Example

Input


2
4 6 5
1 1
1 3
1 5
1 7
1 9
7 10 2
7 8
100 200


Output


0 0 0 2 4 
0 91 
def compress(string):
    string.append("#")
    n = len(string)
    begin, end, cnt = 0, 1, 1
    ans = []
    while end < n:
        if string[begin] == string[end]:
            end, cnt = end + 1, cnt + 1
        else:
            ans.append((string[begin], cnt))
            begin, end, cnt = end, end + 1, 1
    return ans
  
  
def solve(num):
    if num <= init:
        return num
      
    num -= init
    tmp = init
    tmp += (num // diff) * res[2][1]
    num = num - (num // diff) * diff
    tmp += max(num - res[1][1], 0)
    return tmp
    
t = int(input())
for _ in range(t):
    a, b, q = map(int, input().split())
    tmp = [False] * (10 ** 5)
    for i in range(1, 10 ** 5 + 1):
        tmp[i - 1] = (((i) % a) % b == ((i) % b) % a)
    res = compress(tmp)
    
    if len(res) == 1:
        ans = []
        for _ in range(q):
            l, r = map(int, input().split())
            ans.append(0)
        print(*ans) 
        continue
    init = res[0][1]
    diff = res[1][1] + res[2][1]
    ans = []
    # print(init, diff)
    # print(res)
    for _ in range(q):
        l, r = map(int, input().split())
        ans.append((r - solve(r)) - (l - 1 - solve(l - 1)))
    print(*ans)