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 
#include<bits/stdc++.h>
using namespace std;
using ld=long double;
#define s second
#define f first
typedef long long ll ;
int main(){
ll a,b,q,t,s,ss,count=0;
cin>>t;
for(ll i=0;i<t;i++){
cin>>a>>b>>q;
if(a<b)
swap(a,b);
int g=(a*b)/gcd(a,b);
for(ll ii=0;ii<q;ii++){
cin>>s>>ss;
int s1=0,s2=0,ss1=s,ss2=ss,x=0;
count=0;
if(a%b==0 && b!=1){
count=0;
ss2=0;ss1=0;}
else
if(ss<a){
count=0;
ss2=0;ss1=0;}
else{
if(s%g!=0 && s!=0){
s1=s%g;
if(s>g)
s=s+(g-s1);
else
s=9999;
}
if(ss%g!=0 && ss>g){
s2=ss%g;
if(ss>g)
ss=ss-s2;

}
count=(ss-s)/g*(a);
if(s1<a && s1!=0){
count=count+a-s1;
count--;
x++;
}
if(s2<a && s2!=0){
count=count+s2;
if(x==0){
count--;}
}
else{
if(s2!=0)
count=count+a;
if(x==0)
count--;
}

}
cout<<abs((ss2-ss1)-count)<<" ";
}
cout<<endl;
}
 
}