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 namespace std;
long long ans;
int have[10][10], len[10];
int id[10][10];
int g[100010][8];
short f[100010][8];
int h4[100010], h5[100010], h6[100010], t4, t5, t6;
int h1[100010], h2[100010], h3[100010], h[100010], H[100010], num[100010], t1,
    t2, t3, tt;
int to[10][10], t, a, b, c;
bool bo[10][10];
char buf[1000010], *cur = buf + 1000010;
inline char getc() {
  (cur == buf + 1000010) ? fread(cur = buf, 1, 1000010, stdin) : 0;
  return *(cur++);
}
inline long long rd() {
  long long x = 0, f = 1;
  char ch = getc();
  for (; ch < '0' || ch > '9'; ch = getc())
    if (ch == '-') f = -1;
  for (; ch >= '0' && ch <= '9'; ch = getc()) x = x * 10 + ch - '0';
  return x * f;
}
char buff[1000010], *curr = buff;
inline void flush() { fwrite(buff, 1, curr - buff, stdout); }
inline void putc(const char &ch) {
  (curr == buff + 1000010) ? fwrite(curr = buff, 1, 1000010, stdout) : 0;
  *curr++ = ch;
}
inline void print(long long x) {
  static char s[233];
  if (!x) {
    putc('0');
    putc('\n');
    return;
  }
  int tot = 0;
  for (; x; x /= 10) s[++tot] = x % 10 + '0';
  for (; tot; tot--) putc(s[tot]);
  putc('\n');
}
inline void gao2(int *a, int *b, int *c, int t1, int t2, int t3) {
  int now1 = 0, now2 = 0;
  for (int i = 1; i <= t3; i++) {
    while (now1 < t1 && a[now1 + 1] <= c[i]) now1++;
    while (now2 < t2 && b[now2 + 1] <= c[i]) now2++;
    if (a[now1] == c[i] && b[now2] == c[i]) ans++;
  }
}
inline int calc(int x, int y) {
  int res = 0;
  for (register int i = 1; i <= 3; i++)
    for (register int j = i + 1; j <= 3; j++)
      if (((((1 << (i - 1)) & x) && ((1 << (j - 1)) & y)) ||
           (((1 << (i - 1)) & y) && ((1 << (j - 1)) & x))))
        res |= (1 << (id[i][j] - 1));
  return res;
}
inline bool check(int x, int y) {
  if ((x & 1) && (y & 4)) return true;
  if ((x & 2) && (y & 1)) return true;
  if ((x & 4) && (y & 2)) return true;
  return false;
}
int main() {
  t = rd();
  id[1][2] = 1;
  id[2][3] = 2;
  id[1][3] = 3;
  for (int i = 0; i < 8; i++)
    for (int j = 0; j < 8; j++) to[j][i] = calc(i, j), bo[i][j] = check(i, j);
  for (int i = 0; i < 8; i++)
    for (int j = 0; j < 8; j++)
      if (bo[j][i]) have[i][++len[i]] = j;
  while (t--) {
    a = rd();
    b = rd();
    c = rd();
    t1 = t2 = t3 = t4 = t5 = t6 = 0;
    for (int i = 1; i * i <= a; i++)
      if (a % i == 0) {
        h1[++t1] = i;
        if (i * i != a) h4[++t4] = a / i;
      }
    for (int i = 1; i * i <= b; i++)
      if (b % i == 0) {
        h2[++t2] = i;
        if (i * i != b) h5[++t5] = b / i;
      }
    for (int i = 1; i * i <= c; i++)
      if (c % i == 0) {
        h3[++t3] = i;
        if (i * i != c) h6[++t6] = c / i;
      }
    for (int i = t4; i; i--) h1[++t1] = h4[i];
    for (int i = t5; i; i--) h2[++t2] = h5[i];
    for (int i = t6; i; i--) h3[++t3] = h6[i];
    ans = 0;
    gao2(h1, h2, h3, t1, t2, t3);
    tt = 0;
    int now1 = 1, now2 = 1;
    int ttt = 0;
    while (now1 <= tt || now2 <= t1) {
      if (now1 <= tt && now2 <= t1 && h[now1] == h1[now2]) {
        now2++;
        continue;
      }
      if (now1 > tt)
        H[++ttt] = h1[now2++];
      else if (now2 > t1)
        H[++ttt] = h[now1++];
      else if (h[now1] < h1[now2])
        H[++ttt] = h[now1++];
      else
        H[++ttt] = h1[now2++];
    }
    tt = ttt;
    for (int i = 1; i <= tt; i++) h[i] = H[i];
    now1 = 1, now2 = 1;
    ttt = 0;
    while (now1 <= tt || now2 <= t2) {
      if (now1 <= tt && now2 <= t2 && h[now1] == h2[now2]) {
        now2++;
        continue;
      }
      if (now1 > tt)
        H[++ttt] = h2[now2++];
      else if (now2 > t2)
        H[++ttt] = h[now1++];
      else if (h[now1] < h2[now2])
        H[++ttt] = h[now1++];
      else
        H[++ttt] = h2[now2++];
    }
    tt = ttt;
    for (int i = 1; i <= tt; i++) h[i] = H[i];
    now1 = 1, now2 = 1;
    ttt = 0;
    while (now1 <= tt || now2 <= t3) {
      if (now1 <= tt && now2 <= t3 && h[now1] == h3[now2]) {
        now2++;
        continue;
      }
      if (now1 > tt)
        H[++ttt] = h3[now2++];
      else if (now2 > t3)
        H[++ttt] = h[now1++];
      else if (h[now1] < h3[now2])
        H[++ttt] = h[now1++];
      else
        H[++ttt] = h3[now2++];
    }
    tt = ttt;
    for (int i = 1; i <= tt; i++) h[i] = H[i];
    for (int i = 0; i <= tt; i++)
      memset(f[i], 0, sizeof(f[i])), memset(g[i], 0, sizeof(g[i]));
    for (int i = 0; i <= tt; i++) num[i] = 0;
    int now = 0;
    for (int i = 1; i <= t1; i++) {
      while (now < tt && h[now + 1] <= h1[i]) now++;
      num[now] |= 1;
    }
    now = 0;
    for (int i = 1; i <= t2; i++) {
      while (now < tt && h[now + 1] <= h2[i]) now++;
      num[now] |= 2;
    }
    now = 0;
    for (int i = 1; i <= t3; i++) {
      while (now < tt && h[now + 1] <= h3[i]) now++;
      num[now] |= 4;
    }
    for (int i = 1; i <= tt; i++) {
      for (register int j = 1; j <= len[num[i]]; j++)
        ans += g[i - 1][have[num[i]][j]];
      memcpy(f[i], f[i - 1], sizeof(f[i]));
      memcpy(g[i], g[i - 1], sizeof(g[i]));
      for (register int j = 0; j < 8; j++) g[i][to[num[i]][j]] += f[i - 1][j];
      f[i][num[i]]++;
    }
    for (int i = 0; i <= tt; i++) memset(f[i], 0, sizeof(f[i]));
    for (int i = 1; i <= tt; i++) {
      for (register int j = 1; j <= len[num[i]]; j++)
        ans += f[i - 1][have[num[i]][j]];
      memcpy(f[i], f[i - 1], sizeof(f[i]));
      f[i][to[num[i]][num[i]]]++;
    }
    reverse(h + 1, h + tt + 1);
    reverse(num + 1, num + tt + 1);
    for (int i = 0; i <= tt; i++) memset(f[i], 0, sizeof(f[i]));
    for (int i = 1; i <= tt; i++) {
      for (register int j = 1; j <= len[num[i]]; j++)
        ans += f[i - 1][have[num[i]][j]];
      memcpy(f[i], f[i - 1], sizeof(f[i]));
      f[i][to[num[i]][num[i]]]++;
    }
    print(ans);
  }
  flush();
  return 0;
}
