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 lcm(a,b):
    return (a*b)//gcd(a,b)
def gcd(a,b):
    if(a%b==0):
        return b
    else:
        return gcd(b,a%b)
for _ in range(int(input())):
    a,b,q=list(map(int,input().split()))
    if(a<b):
        x=b
        y=a
    else:
        x=a
        y=b
    anslcm=lcm(x,y)
    mul=anslcm-x
    ans=[]
    for w in range(q):
        l,r=list(map(int,input().split()))
        if(a==b):
            ans.append(0)
        else:
            startdiv=l//anslcm
            endiv=r//anslcm
            temp=mul*(endiv-startdiv)
            virstart=startdiv*anslcm
            virend=endiv*anslcm
            tempo=l%anslcm
            if(tempo>x):
                tem=(tempo-x)
                temp-=tem
            tempo=r%anslcm
            if(tempo>x):
                tem=tempo-x+1
                temp+=tem
            ans.append(temp)
    for q in ans:
        print(q,end=' ')
    print()
