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 sys import stdin,stdout
from math import gcd
t=int(stdin.readline().strip())
def f(a,b):
    v=[]
    for i in range(0,100000):
        if (i%a)%b!=(i%b)%a:
            v.append(i)

    print(v)
    return v
for caso in range(t):
    a,b,q=map(int,stdin.readline().strip().split())
    arr=[0 for i in range(max(a,b)+1)]
    v=[]
    g=gcd(a,b)

    if g!=a and g!=b :
        lim=max(a,b)
        for i in range(lim,10**6):
            if (i%a)%b!=(i%b)%a:
                v.append(i)
            else:
                break

    else:
        for i in range(q):
            l,r=map(int,stdin.readline().strip().split())
        ans=[0 for i in range(q)]
        print(*ans)
        continue
    ans=[]
    
    x=v[-1]+1
    for i in range(q):
        l,r=map(int,stdin.readline().strip().split())
        
        l=max(l,v[0])
        y=-((l-1)//x)+(r//x)
        res=y*v[0]
        z=(r//x)*x+v[0]-1
        if((r//x)*x>=l and (r//x)*x<=r):
        
            res-=max(0,z-r)
        u=((l-1)//x)*x
        u1=u+v[0]-1
        if (u1>=l and u1<=r):
            res+=u1-l+1
        if (u<=r and u1>=r):
            res=10**20
        ans.append(max(r-l+1-res,0))
    print(*ans)
