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 
import sys,math
input = sys.stdin.readline

############ ---- USER DEFINED INPUT FUNCTIONS ---- ############
def inp():
    return(int(input()))
def inlt():
    return(list(map(int,input().split())))
def insr():
    s = input()
    return(s[:len(s) - 1])
def invr():
    return(map(int,input().split()))
################################################################
############ ---- THE ACTUAL CODE STARTS BELOW ---- ############

t = inp()
def lcm(a, b):
    return abs(a*b) // math.gcd(a, b)

def find_left(l,lc):
    low,high = 0,10**18 + 1
   
    while low < high:
        mid = (low + high)//2
        if mid*lc>= l:
            high = mid
        else:
            low = mid + 1
        #print(low)
    return low*lc

def find_right(l,lc,r):
    low,high = 1,10**18 + 1
    
    while low < high:
        mid = (low + high)//2
        if l + mid*lc - 1 >= r:
            high = mid
        else:
            low = mid + 1

    return (low,l + low*lc - 1)
for _ in range(t):
    a,b,q = invr()
    ans = []
    ma = max(a,b)
    lc = lcm(a,b)
    for each in range(q):
        l,r = invr()
        left = find_left(l,lc)
        if left > r:
            ol = left - lc
            if ol + ma > r:
                ans.append(0)
            else:
                ans.append(r -  max(ol + ma - 1,l - 1))
        else:
            num,right = find_right(left,lc,r)
            lremain = left - l
            if right > r:
                num -= 1
            lr = min(lremain,lc - ma)
            rr = 0 if right == r else max(0,r - ((right - lc) + ma))
            result = num*(lc - ma) + lr + rr
            ans.append(result)
    print(*ans)
