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).

Source: codeforces | Difficulty: VERY_HARD
Tags: Bit manipulation

Combinatorics

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Number theory

Mathematics


Solutions:
from sys import stdin
from math import gcd

def main():
	input()
	l = stdin.read().splitlines()
	d = [3.0, 1.0, 2.0, 2.0, 2.0, 1.0] * 16667
	for i in range(4, 100001):
		for j in range(i, 100001, i):
			d[j] += 1.0
	for (i, s) in enumerate(l):
		(a, b, c) = map(int, s.split())
		k = gcd(b, c)
		ab = d[gcd(a, b)]
		ac = d[gcd(a, c)]
		bc = d[k]
		abc = d[gcd(a, k)]
		asz = d[a] - ab - ac + abc
		bsz = d[b] - bc - ab + abc
		csz = d[c] - ac - bc + abc
		absz = ab - abc
		bcsz = bc - abc
		acsz = ac - abc
		l[i] = '%d' % (asz * bsz * csz + absz * (asz + bsz) * csz + bcsz * (bsz + csz) * asz + acsz * (asz + csz) * bsz + abc * (asz * bsz + asz * csz + bsz * csz) + abc * (absz + bcsz + acsz) * (asz + bsz + csz) + (asz + bsz + csz + absz + bcsz + acsz) * (abc * (abc + 1) * 0.5) + absz * bcsz * acsz + (absz * (absz + 1.0) * d[c] + bcsz * (bcsz + 1.0) * d[a] + acsz * (acsz + 1.0) * d[b]) * 0.5 + (asz + bsz + csz + abc) * (absz * acsz + absz * bcsz + bcsz * acsz) + (abc + abc * (abc - 1.0) + abc * (abc - 1.0) * (abc - 2.0) / 6.0))
	print('\n'.join(map(str, l)))
main()


N = 100001
fac = [0 for i in range(N)]
for i in range(1, N):
	for j in range(i, N, i):
		fac[j] += 1

def gcd(a, b):
	if a < b:
		(a, b) = (b, a)
	while b > 0:
		(a, b) = (b, a % b)
	return a

def ctt(A, B, C):
	la = fac[A]
	lb = fac[B]
	lc = fac[C]
	ab = gcd(A, B)
	ac = gcd(A, C)
	bc = gcd(B, C)
	abc = gcd(ab, C)
	dupabc = fac[abc]
	dupac = fac[ac] - dupabc
	dupbc = fac[bc] - dupabc
	dupab = fac[ab] - dupabc
	lax = la - dupabc - dupab - dupac
	lbx = lb - dupabc - dupab - dupbc
	lcx = lc - dupabc - dupac - dupbc
	ctx = lax * lbx * lcx
	ctx += lax * lbx * (lc - lcx)
	ctx += lax * lcx * (lb - lbx)
	ctx += lcx * lbx * (la - lax)
	ctx += lax * ((lb - lbx) * (lc - lcx) - (dupabc + dupbc) * (dupabc + dupbc - 1) / 2)
	ctx += lbx * ((la - lax) * (lc - lcx) - (dupabc + dupac) * (dupabc + dupac - 1) / 2)
	ctx += lcx * ((la - lax) * (lb - lbx) - (dupabc + dupab) * (dupabc + dupab - 1) / 2)
	ctx += dupab * dupac * dupbc
	ctx += dupab * dupac * (dupab + dupac + 2) / 2
	ctx += dupab * dupbc * (dupab + dupbc + 2) / 2
	ctx += dupbc * dupac * (dupbc + dupac + 2) / 2
	ctx += dupabc * (dupab * dupac + dupab * dupbc + dupbc * dupac)
	ctx += dupabc * (dupab * (dupab + 1) + (dupbc + 1) * dupbc + (dupac + 1) * dupac) / 2
	ctx += (dupabc + 1) * dupabc * (dupab + dupac + dupbc) / 2
	ctx += dupabc * dupabc + dupabc * (dupabc - 1) * (dupabc - 2) / 6
	return int(ctx)
n = int(input())
for _ in range(n):
	(a, b, c) = map(int, input().split())
	print(ctt(a, b, c))
exit()


from math import gcd
N = 100001
d = [0 for i in range(N)]
for i in range(1, N):
	for j in range(i, N, i):
		d[j] += 1
n = int(input())
for _ in range(n):
	(a, b, c) = map(int, input().split())
	(A, B, C) = (d[a], d[b], d[c])
	(AB, BC, CA) = (d[gcd(a, b)], d[gcd(b, c)], d[gcd(c, a)])
	ABC = d[gcd(gcd(a, b), c)]
	print(A * B * C - AB * BC * CA + ABC * (AB * BC + BC * CA + CA * AB) - A * BC * (BC - 1) // 2 - B * CA * (CA - 1) // 2 - C * AB * (AB - 1) // 2 - ABC * (ABC + 1) * (AB + BC + CA) // 2 + ABC * (ABC + 1) * (ABC + 2) // 6)