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 
#for _ in range(int(input())):
#dt = {} for i in x: dt[i] = dt.get(i,0)+1
#dt = {k:v for k,v in sorted(x.items(), key=lambda i: i[1])}
ipnl = lambda n: [int(input()) for _ in range(n)]
inp = lambda :int(input())
ip = lambda :[int(w) for w in input().split()]
mp = lambda :map(int,input().split())
from math import gcd
for _ in range(int(input())):
    a,b,q = mp()
    flag = 0
    if max(a,b)%min(a,b) == 0:
        flag = 1
    f = []
    for i in range(q):
        l,r = mp()
        if flag: 
            f.append(0)
            continue
        match = 0
        mx = max(a,b)
        p = a*b//gcd(a,b)
        p1 = l//p
        p2 = r//p
        t1 = p*p1
        for j in range(mx):
            if t1 >= l:
                match += 1
            t1 += 1
        t2 = p*p2
        for j in range(mx):
            if t2 <= r:
                match += 1
            t2 += 1
        ans = 0
        match += ((p2-p1)-1)*mx
        tt = r-l+1
        ans = tt - match
        f.append(ans)
    print(*f)
        