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 
import atexit, io, sys 

  
# A stream implementation using an in-memory bytes  
# buffer. It inherits BufferedIOBase. 

buffer = io.BytesIO() 

sys.stdout = buffer

  
# print via here 

@atexit.register 


def write(): 

    sys.__stdout__.write(buffer.getvalue()) 
def gcd(a,b):
    a,b=min(a,b),max(a,b)
    while b:
        a,b=b,a%b
    return a
v=[[0 for j in range(201)]for j in range(201)]
for j in range(1,201):
    for k in range(1,201):
        v[j][k]+=(j*k)/gcd(j,k)

for _ in range(input()):
    
    a,b,q=map(int,raw_input().split())
    fs=0
    for j in range(1,v[a][b]):
        if (j%a)%b!=(j%b)%a:
            fs+=1
    pref=[0]*(v[a][b]+1)
    for j in range(1,v[a][b]+1):
        pref[j]=pref[j-1]
        if (j%a)%b!=(j%b)%a:
            pref[j]+=1
    
    
    for i in range(q):
        l,r=map(int,raw_input().split())
        ans=0
        
        if l%v[a][b]:
            ans+=(fs-pref[l%v[a][b]-1])
        if r%v[a][b]:
            ans+=(pref[r%v[a][b]])
        k=r/v[a][b]
        y=(l-1)/v[a][b]+1
        
        ans+=((k-y)*fs)
        print ans,
    print ""
    