Problem: 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 

Solution (in PYTHON3):

Solution:
import math
t=int(input())
def check1(l,r,aa,bb):
    if (l<=aa and aa<=r and r<=bb) or (aa<=l and l<=bb and bb<=r):
        return True
    return False
for __ in range(t):
    a,b,q=list(map(int,input().split()))
    if a>b:
        a,b=b,a
    gap=b
    inter=(a*b)//math.gcd(a,b)
    #aaaa=[]
    for _ in range(q):
        l,r=list(map(int,input().split()))
        
        #inter*i   l   inter*i+1
        idx1=l//inter
        idx2=r//inter
        #print(l,inter,idx1,idx2,"here")
        #i1,f1=[inter*idx1,inter*idx1+b-1]
        #i2,f2=[inter*idx2,inter*idx2+b-1]
        if inter*idx1+b-1<l:
            idx1+=1
        #i1,f1=[inter*idx1,inter*idx1+b-1]
        
        #i2,f2=[inter*idx2,inter*idx2+b-1]
        #print(idx1,idx2)
        res=0
        if idx2-idx1>6:
            rng=[]
            
            #print(l,r,aa,bb,idx1)
            while idx1>=0:
                #print(idx1)
                aa=inter*idx1
                bb=inter*idx1+b-1
                if check1(l,r,aa,bb):
                    rng.append(idx1)
                else:
                    break
                idx1+=1
            
            while idx2>=0:
                aa=inter*idx2
                bb=inter*idx2+b-1
                if check1(l,r,aa,bb):
                    rng.append(idx2)
                else:
                    break
                idx2-=1
            res=(b)*(idx2-idx1+1)
            #print(len(rng),"here")
            for idx in rng:
                aa=inter*idx
                bb=inter*idx+b-1
                fl=0
                if aa==0:
                    fl=1
                    aa+=1
                
                #print(aa,bb)
                if l>=aa and bb>=r:
                    res+=r-l+1
                elif l<=aa and bb<=r:
                    res+=b-fl
                elif aa<=l and bb>=l:
                    res+=(bb-l+1)
                elif aa<=r and bb>=r:
                    res+=(r-aa+1)


            print(r-l+1-res,end=" ")
            continue

        for idx in range(idx1,idx2+1):
            aa=inter*idx
            bb=inter*idx+b-1
            fl=0
            if aa==0:
                fl=1
                aa+=1
            
            #print(aa,bb)
            if l>=aa and bb>=r:
                res+=r-l+1
            elif l<=aa and bb<=r:
                res+=b-fl
            elif aa<=l and bb>=l:
                res+=(bb-l+1)
            elif aa<=r and bb>=r:
                res+=(r-aa+1)
        print(r-l+1-res,end=" ")
    #print(*aaaa)
    print()

"""
2 3 4 5 6 7


"""