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 gcd(a,b):   
    if (b == 0): 
         return a 
    return gcd(b, a%b)

t=int(input())
while(t):
    t-=1
    a1,b1,q = [int(i) for i in input().split()]
    if a1>b1:
        a1,b1=b1,a1
    ans = []
    while(q):
        a=a1
        b=b1
        q-=1
        l,r = [int(i) for i in input().split()]
        a = a/gcd(a,b)
        count = 0
        start = int(max(l,b)/(a*b))
        end = int(r/(a*b))
        count = (end-start-1)*b
        if r >= (end*a*b) +b:
            count += b
        else:
            count += r-(end*a*b)+1
        if (start*a*b)+b >= max(l,b):
            count += b-(max(l,b)-(start*a*b)) 
        final = r-max(b,l)+1-count
        ans.append(int(final))
    print(*ans)