n = int(input())
s = input().strip()

# Split into three groups based on position mod 3 (1-based)
group1 = []
group2 = []
group3 = []

for i in range(len(s)):
    pos = i + 1  # Convert to 1-based index
    if pos % 3 == 1:
        group1.append(s[i])
    elif pos % 3 == 2:
        group2.append(s[i])
    else:
        group3.append(s[i])

# Count 'c', 'o', 'n' in each group
def count_chars(group):
    c = group.count('c')
    o = group.count('o')
    n = group.count('n')
    return c, o, n

c1, o1, n1 = count_chars(group1)
c2, o2, n2 = count_chars(group2)
c3, o3, n3 = count_chars(group3)

# Calculate the maximum possible 'con' for each triplet type
type1 = min(c1, o2, n3)
type2 = min(c2, o3, n1)
type3 = min(c3, o1, n2)

total = type1 + type2 + type3

print(total)