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 
from math import gcd
for _ in range(int(input())):
    a,b,q = list(map(int,input().split()))
    lcm = (a*b)//(gcd(a,b))
    arr = [0 for i in range(lcm)]
    pref = [0 for i in range(lcm)]
    for i in range(lcm):
        v1 = (i%a)%b
        v2 = (i%b)%a
        if v1!=v2:
            arr[i] = 1
    for i in range(1,lcm):
        pref[i]+=(arr[i]+pref[i-1])
    final = []
    for i in range(q):
        l,r = list(map(int,input().split()))
        x = pref[-1]
        num = r-l+1
        y = l%lcm
        ans = 0
        if(num<=(lcm-y)):
            ans+=(pref[y+num-1]-pref[y])
            if arr[y]:
                ans+=1
            num-=(min(num,lcm-y))
        else:
            num-=(lcm-y)
            ans+=(x-pref[y])
            if arr[y]:
                ans+=1
            ans+=(x*(num//lcm))
            # print('mul',num//lcm,num)
            num = num%lcm
        if num>0:
            ans+=pref[num-1]
        final.append(ans)
    print(*final)