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
from collections import defaultdict as dd
input=sys.stdin.readline
mod=1000000007
def power(x,y): 
    res = 1
    while (y > 0):  
        if ((y & 1) == 1) : 
            res = (res * x )%m
        y = y >> 1
        x =( x * x )%m
      
    return res%m
def gcd(a,b): 
    if a == 0: 
        return b 
    return gcd(b % a, a)
def solve(r,mx,lc):
    mulp=0
    ul=int(r//lc)*lc
    le=mx-1
##            fm=(int(l//lc))*lc
##            le=max(fm+mx-l,0)
            
            
##            if(l%lc==0):
##                le=0
    if(ul+mx>r):
        le+=r-ul+1
    else:
        le+=mx
##            print(le)    
##            if(r%lc==0):
##                mulp+=1
            
    mulp=int(r//lc)-1
    le+=mulp*mx
    return(r-le+1)

def lcm(a,b): 
    return (a*b)//gcd(a,b)
for _ in range(int(input())):   
    a,b,q=map(int,input().split())
    lc=lcm(a,b)
    mx=max(a,b)
    f=0
    if(mx%min(a,b)==0):
        f=1
    for gf in range(q):
        c=0
        l,r=map(int,input().split())
        if(mx%min(a,b)==0):
            ans=0
        elif(r<mx):
            ans=0
        else:
            su1=solve(r,mx,lc)
            su2=solve(l-1,mx,lc)
##            print(su1-su2,su1,su2)
            ans=su1-su2
        print(ans,end=" ")
    print()
        
##            for i in range(l,r+1):
##                x=(i%a)%b;y=(i%b)%a
##                if(x != y ):
##                    c+=1
####'''
####6 15 1
####1 100
####45
####4 5 1
####1 1000
####750
####4 6 1
####100 200
####51'''
######                    print(i,x,y)
######                else:
######                    print(i)
##            print("final",c)
##            print(r-l+1-le,le)
##    if(f==0):
##         print(*li)
##            print(r-l+1-le)
            
##            le=r-ul+1
##            le=le+max((mulp-1)*mx,0)
##            c+=fm-l+1
##            le+=max(
            

##    print(min((abs(x-y)*a + min(x,y)*b),(x+y)*a))
