题目链接：http://codeforces.com/contest/1008/problem/D
思路：要铺满明显只要a，b，c分别是A，B，C的因数就好了，但a，b，c可能有重复的排列出现要去掉。。我自己是画了张图列举出了所以情况。。应该有简单的办法吧，但是想不到emmm。
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <bitset>
#include <cmath>
#include <cctype>
#include <iostream>
#include <algorithm>
#include <string>
#include <vector>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <sstream>
#include <iomanip>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
const ll inff = 0x3f3f3f3f3f3f3f3f;
#define FOR(i,a,b) for(int i(a);i<=(b);++i)
#define FOL(i,a,b) for(int i(a);i>=(b);--i)
#define REW(a,b) memset(a,b,sizeof(a))
#define inf int(0x3f3f3f3f)
#define si(a) scanf("%d",&a)
#define sl(a) scanf("%I64d",&a)
#define sd(a) scanf("%lf",&a)
#define ss(a) scanf("%s",a)
#define mod int(1e9+7)
#define pb push_back
#define lc (d<<1)
#define Pll pair<ll,ll>
#define P pair<int,int>
#define pi acos(-1)
int t,x,y,z,ds[100008],a,b,c,ab,ac,bc,abc;
int main()
{
cin.tie(0);
cout.tie(0);
cin>>t;
FOR(i,1,100000)
{
for(int j=i;j<=100000;j+=i) ds[j]++;
}
while(t--)
{
int ans=0;
si(x),si(y),si(z);
abc=__gcd(x,y),abc=__gcd(abc,z),abc=ds[abc];
a=ds[x]-ds[__gcd(x,y)]-ds[__gcd(x,z)]+abc;
b=ds[y]-ds[__gcd(x,y)]-ds[__gcd(y,z)]+abc;
c=ds[z]-ds[__gcd(x,z)]-ds[__gcd(y,z)]+abc;
ab=ds[__gcd(x,y)]-abc;
ac=ds[__gcd(x,z)]-abc;
bc=ds[__gcd(z,y)]-abc;
if(a) ans+=a*(ab+abc+bc)*(bc+abc+ac)-(abc+bc)*(abc+bc-1)*a/2;
if(b) ans+=b*(ab+abc+ac)*(bc+abc+ac)-(abc+ac)*(abc+ac-1)*b/2;
if(c) ans+=c*(ac+abc+ab)*(bc+abc+ab)-(abc+ab)*(abc+ab-1)*c/2;
ans+=a*b*(bc+abc+ac);
ans+=b*c*(ab+abc+ac);
ans+=a*c*(bc+abc+ab);
ans+=a*b*c;
ans+=(abc-1)*(abc-2)*abc/6+(abc-1)*abc+abc;
ans+=(ab+bc+ac)*2*abc;
ans+=(abc-1)*abc*(ab+bc+ac)/2;
ans+=abc*(ab+bc+ac-1)*(ab+ac+bc)/2;
ans+=ab*(ac+bc)+ac*(ab+bc)+bc*(ab+ac);
ans+=(ab+ac+bc-2)*(ab+ac+bc-1)*(ab+ac+bc)/6;
ans-=(ab-2)*(ab-1)*ab/6;
ans-=(ac-2)*(ac-1)*ac/6;
ans-=(bc-2)*(bc-1)*bc/6;
cout<<ans<<endl;
}
return 0;
}