# 2文字目を固定すると、（自分より前の同じ文字の個数）*(自分より後ろの自分以外の文字の数)
# というわけで、現時点の自分と同じ文字のカウントがあれば良さそう


S = input()
n = len(S)

alphabet = "ABCDEFGHIJKLMNOPQRSTUVWXYZ"
# assert len(alphabet) == len(set(list(alphabet))) == 26
ci_map = dict()

for i, c in enumerate(alphabet):
	ci_map[c] = i

# 自分より前の同じ文字の個数
counter = [0] * 26
same_before = [0] * n
for i, c in enumerate(S):
	j = ci_map[c]
	same_before[i] = counter[j]
	counter[j] += 1

res = 0
for i in range(n):
	before = same_before[i]
	c = S[i]
	j = ci_map[c]
	after = n - 1 - i - (counter[j] - before - 1)
	res += before * after

print(res)