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 


def gcd(a,b): 
    if a == 0: 
        return b 
    return gcd(b % a, a) 
  

def compute_lcm(a,b): 
    return (a*b) // gcd(a,b) 
def find(r,a,b):
    lcm=compute_lcm(a,b)
    t=((r//(lcm))+1)
    if r%(lcm)<max(a,b):
        t-=1

        ans=((r//(lcm))*(lcm)-t*max(a,b))
    else:
        ans=(r-t*max(a,b))+1
    return ans
n=int(input())
for _ in range(n):
    a,b,q=map(int,input().split())
    final=[]
    for k in range(q):
        l,r=map(int,input().split())
        if a>b and a%b==0:
            final.append(0)
        elif a<=b and b%a==0:
            final.append(0)
        else:
            final.append(find(r,a,b)-find(l-1,a,b))
    print(" ".join(map(str,final)))


##a,b=4,6
##for l in range(1,1000):
##    for r in range(l+1,1000):
##        #r=100
##        count=0
##        for i in range(l,r+1):
##            if (i%a)%b!=(i%b)%a:
##                #print(i)
##                count+=1
##    
##        ansA=(count)
##        ans=find(r,a,b)-find(l-1,a,b)
##        #print(ans,ansA)
##        if(ansA!=ans):
##            print(l,r,ansA,ans)
##count=0
##l,r=0,158
##for i in range(l,r+1):
##    if (i%a)%b!=(i%b)%a:
##        print(i)
##        count+=1
##    else:
##        print("sad " +str(i))
##print(count,find(r,a,b)-find(l-1,a,b))
