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,ceil,floor
import io,os
input = io.BytesIO(os.read(0,os.fstat(0).st_size)).readline
def lcm(a, b):
    return abs(a*b) // gcd(a, b)
for _ in range(int(input())):
    a,b,q=map(int,input().split())
    l=[]
    for i in range(q):
        l.append(list(map(int,input().split())))
    fr1,fr2=lcm(a,b),max(a,b)
    for s,e in l:
        ans=0
        t1=(s//fr1 +(1 if s % fr1 != 0 else 0))*fr1
        t2=t1-fr1
        if t1 < e:
            kk=max(s,fr2+t2)
            ans+=t1-kk
            t3=(e//fr1 + (1 if e % fr1 != 0 else 0 ))*fr1
            t4=t3-fr1
            if e % fr1 == 0:
                e-=1
            if fr2 <= 0:
                ans+=1
            tt=e-(fr2+t4)
            if tt >=0:
                ans+=tt+1
            else:
                pass
            ans+=((t4-t1)//fr1)*(fr1-fr2)
            print(ans,end=" ")
        else:
            if e % fr1 == 0:
                e-=1
            if fr2 <= 0:
                ans+=1
            kk=max(s,fr2+t2)
            kk2=min(e,t1-1)
            ans+=max(kk2-kk+1,0)
            print(ans,end=" ")
    print()