You are given a rectangular parallelepiped with sides of positive integer lengths A, B and C. 

Find the number of different groups of three integers (a, b, c) such that 1≤ a≤ b≤ c and parallelepiped A× B× C can be paved with parallelepipeds a× b× c. Note, that all small parallelepipeds have to be rotated in the same direction.

For example, parallelepiped 1× 5× 6 can be divided into parallelepipeds 1× 3× 5, but can not be divided into parallelepipeds 1× 2× 3.

Input

The first line contains a single integer t (1 ≤ t ≤ 10^5) — the number of test cases.

Each of the next t lines contains three integers A, B and C (1 ≤ A, B, C ≤ 10^5) — the sizes of the parallelepiped.

Output

For each test case, print the number of different groups of three points that satisfy all given conditions.

Example

Input

4
1 1 1
1 6 1
2 2 2
100 100 100


Output

1
4
4
165

Note

In the first test case, rectangular parallelepiped (1, 1, 1) can be only divided into rectangular parallelepiped with sizes (1, 1, 1).

In the second test case, rectangular parallelepiped (1, 6, 1) can be divided into rectangular parallelepipeds with sizes (1, 1, 1), (1, 1, 2), (1, 1, 3) and (1, 1, 6).

In the third test case, rectangular parallelepiped (2, 2, 2) can be divided into rectangular parallelepipeds with sizes (1, 1, 1), (1, 1, 2), (1, 2, 2) and (2, 2, 2). 
//#pragma GCC target("avx2")
//#pragma GCC optimize("O3")

//# include <x86intrin.h>
# include <bits/stdc++.h>

# include <ext/pb_ds/assoc_container.hpp>
# include <ext/pb_ds/tree_policy.hpp>

using namespace __gnu_pbds;
using namespace std;
 
template<typename T> using ordered_set = tree <T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;

#define _USE_MATH_DEFINES_
#define ll long long
#define ld long double
#define Accepted 0
#define pb push_back
#define mp make_pair
#define sz(x) (int)(x.size())
#define every(x) x.begin(),x.end()
#define F first
#define S second
#define lb lower_bound
#define ub upper_bound
#define For(i,x,y)  for (ll i = x; i <= y; i ++) 
#define FOr(i,x,y)  for (ll i = x; i >= y; i --)
#define SpeedForce ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0)
// ROAD to...                                                                                                                                                                                                                Red

inline void Input_Output () {
	//freopen(".in", "r", stdin);
   //freopen(".out", "w", stdout);
}

const double eps = 0.000001;
const ld pi = acos(-1);
const int maxn = 1e7 + 9;
const int mod = 1e9 + 7;
const ll MOD = 1e18 + 9;
const ll INF = 1e18 + 123;
const int inf = 2e9 + 11;
const int mxn = 1e6 + 9;
const int N = 6e5 + 123;                                          
const int M = 22;
const int pri = 997;
const int Magic = 2101;

const int dx[] = {-1, 0, 1, 0};
const int dy[] = {0, -1, 0, 1};
 
int n, m, k;
vector < int > g[N];
ll f[N];
void solve () {
	int a, b, c;
	cin >> a >> b >> c;
	if (a > b) swap(a, b);
	if (b > c) swap(b, c);
	if (a > b) swap(a, b);

	ll ans = 0;
	/*for (int ic = 0; ic < sz(g[c]); ++ic) {
		for (int ib = 0; ib < sz(g[b]) && g[b][ib] <= g[c][ic]; ++ib) {
			for (int ia = 0; ia < sz(g[a]) && g[a][ia] <= g[b][ib]; ++ia) {
				++ans;
			}
		}
	}*/

	f[0] = 0;
	for (int ib = 0, ia = 0; ib < sz(g[b]); ++ib) {
		while(ia < sz(g[a]) && g[a][ia] <= g[b][ib]) {
			++ia;
		}
		f[ib] += ia;
		f[ib+1] = f[ib];
	}
	for (int ic = 0, ib = 0; ic < sz(g[c]); ++ic) {
		while(ib < sz(g[b]) && g[b][ib] <= g[c][ic]) {
			++ib;
		}
		//cout << ib << ' ' << f[ib] << '\n';

		ans += f[ib-1];
	}


	cout << ans << '\n';
}


int main () {
   	SpeedForce;
   	for (int i = 1; i <= 1e5; ++i)
   		for (int j = i; j <= 1e5; j += i)
   			g[j].pb(i);
	int tt;
	cin >> tt;
	while(tt--) solve();

   	return Accepted;
}

// B...a