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
input = sys.stdin.buffer.readline  
t=int(input())
from math import gcd
while t:
    t-=1
    a,b,q=map(int,input().split())
    lcm=(a*b)//gcd(a,b)
    ans=[None]*q
    if a>b: a,b=b,a
    if gcd(a,b)==a:
        for _ in range(q):
           l,r=map(int,input().split())
        print(*[0]*q)
    else:
        for _ in range(q):
            l,r=map(int,input().split())
            d=0
            last=min(r+1,((l-1)//lcm+1)*lcm)
            for i in range(l,last):
                if (i%a)%b==(i%b)%a:
                    d+=1
                else: break
            if last==r+1:
                ans[_]=(r-l+1-d)
                continue
            nl=(l-1)//lcm
            nr=r//lcm
            nb=nr-nl
            d+=max(0,(nb-1)*b)+min(b,r-nr*lcm+1)
            ans[_]=r-l+1-d
        print(*ans)