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
t=input()
for _ in range(int(t)):
    a,b,q=map(int,input().split())
    lcm=a*b//gcd(a,b)
    for __ in range(q):
        l,r=map(int,input().split())
        common=(r//lcm)-((l-1)//lcm)
        if common==0 and (l%a)%b!=(l%b)%a and (r%a)%b!=(r%b)%a:
            ans=r-l+1
        else:
            noteq=r-l+1
            eq=0
            if r%a==0 and r%b==0:
                common-=1
                eq+=1
            if (r%a)%b==(r%b)%a and not(r%a==0 and r%b==0):
                if (r%a)%b==0 and r%min(a,b)==0:
                    eq+=min(r-l+1,min(a,b)+1)
                else:
                    eq+=min(r-l+1,((r%a)%b)+1)
            if (l%a)%b==(l%b)%a:
                c=0
                for i in range(l,r+1):
                    if (i%a)%b!=(i%b)%a:
                        break
                    c+=1
                eq+=min(r-l+1,c)
            eq+=(max(0,common)*max(a,b))
            noteq=noteq-eq
            ans=max(0,noteq)
        print(ans,end=' ')
    print('')