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 cal(x,y):
    #最小公倍数
    a,b = x,y
    while b>0:
        a,b = b,a%b
    return int(a*((x/a)*(y/a)))
def do(num,M,maxx):
    ans = 0
    if (num%M)>maxx:ans += (num%M)-maxx
    ans += (num//M)*(M-maxx)
    return int(ans)

N = int(input())
for _ in range(N):
    a,b,q = list(map(int,input().split(' ')))
    M = cal(a,b)
    tmp = []
    for i in range(q):
        l,r = list(map(int,input().split(' ')))
        res = do(r+1,M,max(a,b))-do(l,M,max(a,b))
        tmp.append(res)
    for i in range(len(tmp)-1):
        print(tmp[i],end=' ')
    print(tmp[len(tmp)-1])
