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). 
#include <bits/stdc++.h>
using ll = long long;
using namespace std;
template <typename T>
ostream& operator<<(ostream& out, vector<T> const& xs) {
  for (int i = 0; (i) < int(int(xs.size()) - 1); ++(i)) out << xs[i] << ' ';
  if (not xs.empty()) out << xs.back();
  return out;
}
ll powmod(ll x, ll y, ll m) {
  assert(0 <= x and x < m);
  assert(0 <= y);
  ll z = 1;
  for (ll i = 1; i <= y; i <<= 1) {
    if (y & i) z = z * x % m;
    x = x * x % m;
  }
  return z;
}
ll modinv(ll x, ll p) {
  assert(x % p != 0);
  return powmod(x, p - 2, p);
}
template <int32_t MOD>
int32_t fact(int n) {
  static vector<int32_t> memo(1, 1);
  while (n >= memo.size()) {
    memo.push_back(memo.back() * (int64_t)memo.size() % MOD);
  }
  return memo[n];
}
template <int32_t PRIME>
int32_t inv_fact(int n) {
  static vector<int32_t> memo(1, 1);
  while (n >= memo.size()) {
    memo.push_back(memo.back() * (int64_t)modinv(memo.size(), PRIME) % PRIME);
  }
  return memo[n];
}
template <int MOD>
int choose(int n, int r) {
  assert(0 <= r and r <= n);
  return fact<MOD>(n) * (ll)inv_fact<MOD>(n - r) % MOD * (ll)inv_fact<MOD>(r) %
         MOD;
}
template <int MOD>
int multichoose(int n, int r) {
  if (n == 0 and r == 0) return 1;
  return choose<MOD>(n + r - 1, r);
}
void split_to_disjoint(const vector<int>& a, const vector<int>& b,
                       const vector<int>& c, array<vector<int>, 8>& d) {
  int i = 0, j = 0, k = 0;
  while (a[i] != INT_MAX or b[j] != INT_MAX or c[k] != INT_MAX) {
    int x = min({a[i], b[j], c[k]});
    int mask = (x == a[i] ? 1 : 0) | (x == b[j] ? 2 : 0) | (x == c[k] ? 4 : 0);
    d[mask].push_back(x);
    if (x == a[i]) ++i;
    if (x == b[j]) ++j;
    if (x == c[k]) ++k;
  }
}
ll count_increasing(int i, int j, int k, const array<vector<int>, 8>& d) {
  constexpr int MOD = 1000000007;
  int a = d[i].size();
  int b = d[j].size();
  int c = d[k].size();
  if (not a or not b or not c) return 0;
  if (i == j and j == k) {
    return multichoose<MOD>(a, 3);
  } else if (i == j) {
    return multichoose<MOD>(a, 2) * c;
  } else if (j == k) {
    return multichoose<MOD>(b, 2) * a;
  } else if (k == i) {
    return multichoose<MOD>(c, 2) * b;
  } else {
    return a * b * c;
  }
}
int main() {
  constexpr int LIMIT = 100000;
  vector<vector<int> > factors(LIMIT + 1);
  for (int x = (1); (x) < int(LIMIT + 1); ++(x)) {
    for (int kx = x; kx <= LIMIT; kx += x) {
      factors[kx].push_back(x);
    }
  }
  for (int x = 0; (x) < int(LIMIT + 1); ++(x)) {
    factors[x].push_back(INT_MAX);
  }
  vector<array<int, 3> > ps;
  for (int i : {0x1, 0x3, 0x5, 0x7}) {
    for (int j : {0x2, 0x3, 0x6, 0x7}) {
      for (int k : {0x4, 0x5, 0x6, 0x7}) {
        array<int, 3> p = {{i, j, k}};
        sort(begin(p), end(p));
        ps.push_back(p);
      }
    }
  }
  sort(begin(ps), end(ps));
  ps.erase(unique(begin(ps), end(ps)), ps.end());
  int t;
  scanf("%d", &t);
  vector<int> a(t), b(t), c(t);
  for (int i = 0; (i) < int(t); ++(i)) {
    scanf("%d%d%d", &a[i], &b[i], &c[i]);
  }
  vector<ll> cnt(t);
  for (int i = 0; (i) < int(t); ++(i)) {
    array<vector<int>, 8> d;
    split_to_disjoint(factors[a[i]], factors[b[i]], factors[c[i]], d);
    for (auto p : ps) {
      cnt[i] += count_increasing(p[0], p[1], p[2], d);
    }
  }
  for (int i = 0; (i) < int(t); ++(i)) {
    printf("%lld\n", cnt[i]);
  }
  return 0;
}
