Buggy python submission (verdict: Time Limit Exceeded):
def li():
    return list(map(int, input().split()))


if __name__ == "__main__":
    s = input()
    ope_num = 0
    index = 0
    a_stock = 0
    b_stock = 0
    s = list(s)
    while index < len(s):
        # print(s)
        # print(index)
        # print(ope_num)
        # print()
        si = s[index]
        if si == "A":
            if b_stock > 0:
                a_stock = 1
            else:
                a_stock = a_stock + 1
            b_stock = 0
        elif a_stock > 0:
            if b_stock == 0:
                if si == "B":
                    b_stock = 1
                else:
                    a_stock = 0
                    b_stock = 0
            elif si == "C":
                ope_num = ope_num + a_stock
                # s[index - 2 - a_stock + 1] = "B"
                # s[index - 1 - a_stock + 1] = "C"
                for i in range(a_stock - 1, -1, -1):
                    s[index - i] = "A"
                index = index - a_stock
                a_stock = 0
                b_stock = 0
            else:
                a_stock = 0
                b_stock = 0
        else:
            a_stock = 0
            b_stock = 0
        index = index + 1
    print(ope_num)

# ABCの数
# ABCの前にあるAの分だけ追加で操作できる
# ABCの後にある(BC)の分だけ追加で操作できる

# Aがあれば覚えておく
# ABCと来たら、それより前側に隣接しているAの回数だけ追加で操作できる
# 見つかったABCをBCAに置き換える
# BCAのAから走査を継続する


Fixed submission:
def li():
    return list(map(int, input().split()))


if __name__ == "__main__":
    s = input()
    ope_num = 0
    index = 0
    a_stock = 0
    b_stock = 0
    s = list(s)
    while index < len(s):
        # print(s)
        # print(index)
        # print(ope_num)
        # print()
        si = s[index]
        if si == "A":
            if b_stock > 0:
                a_stock = 1
            else:
                a_stock = a_stock + 1
            b_stock = 0
        elif a_stock > 0:
            if b_stock == 0:
                if si == "B":
                    b_stock = 1
                else:
                    a_stock = 0
                    b_stock = 0
            elif si == "C":
                ope_num = ope_num + a_stock
                # s[index - 2 - a_stock + 1] = "B"
                # s[index - 1 - a_stock + 1] = "C"
                # s[index] = "A"
                # index = index - 1
                a_stock = a_stock
                b_stock = 0
            else:
                a_stock = 0
                b_stock = 0
        else:
            a_stock = 0
            b_stock = 0
        index = index + 1
    print(ope_num)

# ABCの数
# ABCの前にあるAの分だけ追加で操作できる
# ABCの後にある(BC)の分だけ追加で操作できる

# Aがあれば覚えておく
# ABCと来たら、それより前側に隣接しているAの回数だけ追加で操作できる
# 見つかったABCをBCAに置き換える
# BCAのAから走査を継続する

Buggy python submission (verdict: Time Limit Exceeded):
def li():
    return list(map(int, input().split()))


if __name__ == "__main__":
    s = input()
    ope_num = 0
    index = 0
    a_stock = 0
    b_stock = 0
    s = list(s)
    while index < len(s):
        # print(s)
        # print(index)
        # print(ope_num)
        # print()
        si = s[index]
        if si == "A":
            if b_stock > 0:
                a_stock = 1
            else:
                a_stock = a_stock + 1
            b_stock = 0
        elif a_stock > 0:
            if b_stock == 0 and si == "B":
                b_stock = 1
            elif b_stock > 0 and si == "C":
                ope_num = ope_num + a_stock
                # s[index - 2 - a_stock + 1] = "B"
                # s[index - 1 - a_stock + 1] = "C"
                for i in range(a_stock - 1, -1, -1):
                    s[index - i] = "A"
                index = index - a_stock
                a_stock = 0
                b_stock = 0
            else:
                a_stock = 0
                b_stock = 0
        else:
            a_stock = 0
            b_stock = 0
        index = index + 1
    print(ope_num)

# ABCの数
# ABCの前にあるAの分だけ追加で操作できる
# ABCの後にある(BC)の分だけ追加で操作できる

# Aがあれば覚えておく
# ABCと来たら、それより前側に隣接しているAの回数だけ追加で操作できる
# 見つかったABCをBCAに置き換える
# BCAのAから走査を継続する


Fixed submission:
def li():
    return list(map(int, input().split()))


if __name__ == "__main__":
    s = input()
    ope_num = 0
    index = 0
    a_stock = 0
    b_stock = 0
    s = list(s)
    while index < len(s):
        # print(s)
        # print(index)
        # print(ope_num)
        # print()
        si = s[index]
        if si == "A":
            if b_stock > 0:
                a_stock = 1
            else:
                a_stock = a_stock + 1
            b_stock = 0
        elif a_stock > 0:
            if b_stock == 0:
                if si == "B":
                    b_stock = 1
                else:
                    a_stock = 0
                    b_stock = 0
            elif si == "C":
                ope_num = ope_num + a_stock
                # s[index - 2 - a_stock + 1] = "B"
                # s[index - 1 - a_stock + 1] = "C"
                # s[index] = "A"
                # index = index - 1
                a_stock = a_stock
                b_stock = 0
            else:
                a_stock = 0
                b_stock = 0
        else:
            a_stock = 0
            b_stock = 0
        index = index + 1
    print(ope_num)

# ABCの数
# ABCの前にあるAの分だけ追加で操作できる
# ABCの後にある(BC)の分だけ追加で操作できる

# Aがあれば覚えておく
# ABCと来たら、それより前側に隣接しているAの回数だけ追加で操作できる
# 見つかったABCをBCAに置き換える
# BCAのAから走査を継続する

Buggy python submission (verdict: Time Limit Exceeded):
def li():
    return list(map(int, input().split()))


if __name__ == "__main__":
    s = input()
    ope_num = 0
    index = 0
    a_stock = 0
    b_stock = 0
    s = list(s)
    while index < len(s):
        # print(s)
        # print(index)
        # print(ope_num)
        # print()
        si = s[index]
        if si == "A":
            if b_stock > 0:
                a_stock = 1
            else:
                a_stock = a_stock + 1
            b_stock = 0
        elif a_stock > 0:
            if si == "B" and b_stock == 0:
                b_stock = 1
            elif si == "C" and b_stock > 0:
                ope_num = ope_num + a_stock
                s[index - 2 - a_stock + 1] = "B"
                s[index - 1 - a_stock + 1] = "C"
                for i in range(a_stock - 1, -1, -1):
                    s[index - i] = "A"
                index = index - a_stock
                a_stock = 0
                b_stock = 0
            else:
                a_stock = 0
                b_stock = 0
        else:
            a_stock = 0
            b_stock = 0
        index = index + 1
    print(ope_num)

# ABCの数
# ABCの前にあるAの分だけ追加で操作できる
# ABCの後にある(BC)の分だけ追加で操作できる

# Aがあれば覚えておく
# ABCと来たら、それより前側に隣接しているAの回数だけ追加で操作できる
# 見つかったABCをBCAに置き換える
# BCAのAから走査を継続する


Fixed submission:
def li():
    return list(map(int, input().split()))


if __name__ == "__main__":
    s = input()
    ope_num = 0
    index = 0
    a_stock = 0
    b_stock = 0
    s = list(s)
    while index < len(s):
        # print(s)
        # print(index)
        # print(ope_num)
        # print()
        si = s[index]
        if si == "A":
            if b_stock > 0:
                a_stock = 1
            else:
                a_stock = a_stock + 1
            b_stock = 0
        elif a_stock > 0:
            if b_stock == 0:
                if si == "B":
                    b_stock = 1
                else:
                    a_stock = 0
                    b_stock = 0
            elif si == "C":
                ope_num = ope_num + a_stock
                # s[index - 2 - a_stock + 1] = "B"
                # s[index - 1 - a_stock + 1] = "C"
                # s[index] = "A"
                # index = index - 1
                a_stock = a_stock
                b_stock = 0
            else:
                a_stock = 0
                b_stock = 0
        else:
            a_stock = 0
            b_stock = 0
        index = index + 1
    print(ope_num)

# ABCの数
# ABCの前にあるAの分だけ追加で操作できる
# ABCの後にある(BC)の分だけ追加で操作できる

# Aがあれば覚えておく
# ABCと来たら、それより前側に隣接しているAの回数だけ追加で操作できる
# 見つかったABCをBCAに置き換える
# BCAのAから走査を継続する

Buggy python submission (verdict: Time Limit Exceeded):
def li():
    return list(map(int, input().split()))


if __name__ == "__main__":
    s = input()
    ope_num = 0
    index = 0
    a_stock = 0
    b_stock = 0
    s = list(s)
    while index < len(s):
        si = s[index]
        # print(index)
        # print(si)
        # print(a_stock)
        # print(b_stock)
        # print(ope_num)
        # print()
        if si == "A":
            b_stock = 0
            a_stock = 1
        elif a_stock > 0:
            if si == "B" and b_stock == 0:
                b_stock = 1
            elif si == "C" and b_stock > 0:
                ope_num = ope_num + 1
                s[index - 2] = "B"
                s[index - 1] = "C"
                s[index] = "A"
                a_stock = 0
                b_stock = 0
                if index - 4 >= -1:
                    index = index - 4
                else:
                    index = index - 1
            else:
                a_stock = 0
                b_stock = 0
        else:
            a_stock = 0
            b_stock = 0
        index = index + 1
    print(ope_num)

# ABCの数
# ABCの前にあるAの分だけ追加で操作できる
# ABCの後にある(BC)の分だけ追加で操作できる

# Aがあれば覚えておく
# ABCと来たら、それより前側に隣接しているAの回数だけ追加で操作できる
# 見つかったABCをBCAに置き換える
# BCAのAから走査を継続する


Fixed submission:
def li():
    return list(map(int, input().split()))


if __name__ == "__main__":
    s = input()
    ope_num = 0
    index = 0
    a_stock = 0
    b_stock = 0
    s = list(s)
    while index < len(s):
        # print(s)
        # print(index)
        # print(ope_num)
        # print()
        si = s[index]
        if si == "A":
            if b_stock > 0:
                a_stock = 1
            else:
                a_stock = a_stock + 1
            b_stock = 0
        elif a_stock > 0:
            if b_stock == 0:
                if si == "B":
                    b_stock = 1
                else:
                    a_stock = 0
                    b_stock = 0
            elif si == "C":
                ope_num = ope_num + a_stock
                # s[index - 2 - a_stock + 1] = "B"
                # s[index - 1 - a_stock + 1] = "C"
                # s[index] = "A"
                # index = index - 1
                a_stock = a_stock
                b_stock = 0
            else:
                a_stock = 0
                b_stock = 0
        else:
            a_stock = 0
            b_stock = 0
        index = index + 1
    print(ope_num)

# ABCの数
# ABCの前にあるAの分だけ追加で操作できる
# ABCの後にある(BC)の分だけ追加で操作できる

# Aがあれば覚えておく
# ABCと来たら、それより前側に隣接しているAの回数だけ追加で操作できる
# 見つかったABCをBCAに置き換える
# BCAのAから走査を継続する

Buggy python submission (verdict: Time Limit Exceeded):
def li():
    return list(map(int, input().split()))


if __name__ == "__main__":
    s = input()
    ope_num = 0
    index = 0
    a_stock = 0
    b_stock = 0
    s = list(s)
    while index < len(s):
        si = s[index]
        # print(index)
        # print(si)
        # print(a_stock)
        # print(b_stock)
        # print(ope_num)
        # print()
        if si == "A":
            if b_stock > 0:
                b_stock = 0
                a_stock = 1
            else:
                a_stock = 1
        elif si == "B" and a_stock > 0 and b_stock == 0:
            b_stock = 1
        elif si == "C" and a_stock > 0 and b_stock > 0:
            ope_num = ope_num + 1
            s[index - 2] = "B"
            s[index - 1] = "C"
            s[index] = "A"
            a_stock = 0
            b_stock = 0
            if index - 4 >= -1:
                index = index - 4
            else:
                index = index - 1
        else:
            a_stock = 0
            b_stock = 0
        index = index + 1
    print(ope_num)

# ABCの数
# ABCの前にあるAの分だけ追加で操作できる
# ABCの後にある(BC)の分だけ追加で操作できる

# Aがあれば覚えておく
# ABCと来たら、それより前側に隣接しているAの回数だけ追加で操作できる
# 見つかったABCをBCAに置き換える
# BCAのAから走査を継続する


Fixed submission:
def li():
    return list(map(int, input().split()))


if __name__ == "__main__":
    s = input()
    ope_num = 0
    index = 0
    a_stock = 0
    b_stock = 0
    s = list(s)
    while index < len(s):
        # print(s)
        # print(index)
        # print(ope_num)
        # print()
        si = s[index]
        if si == "A":
            if b_stock > 0:
                a_stock = 1
            else:
                a_stock = a_stock + 1
            b_stock = 0
        elif a_stock > 0:
            if b_stock == 0:
                if si == "B":
                    b_stock = 1
                else:
                    a_stock = 0
                    b_stock = 0
            elif si == "C":
                ope_num = ope_num + a_stock
                # s[index - 2 - a_stock + 1] = "B"
                # s[index - 1 - a_stock + 1] = "C"
                # s[index] = "A"
                # index = index - 1
                a_stock = a_stock
                b_stock = 0
            else:
                a_stock = 0
                b_stock = 0
        else:
            a_stock = 0
            b_stock = 0
        index = index + 1
    print(ope_num)

# ABCの数
# ABCの前にあるAの分だけ追加で操作できる
# ABCの後にある(BC)の分だけ追加で操作できる

# Aがあれば覚えておく
# ABCと来たら、それより前側に隣接しているAの回数だけ追加で操作できる
# 見つかったABCをBCAに置き換える
# BCAのAから走査を継続する

Buggy python submission (verdict: Runtime Error):
def li():
    return list(map(int, input().split()))


if __name__ == "__main__":
    s = input()
    ope_num = 0
    index = 0
    a_stock = 0
    b_stock = 0
    s = list(s)
    while index < len(s):
        si = s[index]
        # print(index)
        # print(si)
        # print(a_stock)
        # print(b_stock)
        # print(ope_num)
        # print()
        if si == "A":
            if b_stock > 0:
                b_stock = 0
                a_stock = 1
            else:
                a_stock = 1
        elif si == "B" and a_stock > 0 and b_stock == 0:
            b_stock.append(si)
        elif si == "C" and a_stock > 0 and b_stock > 0:
            ope_num = ope_num + 1
            s[index - 2] = "B"
            s[index - 1] = "C"
            s[index] = "A"
            a_stock = 0
            b_stock = 0
            if index - 4 >= -1:
                index = index - 4
            else:
                index = index - 1
        else:
            a_stock = 0
            b_stock = 0
        index = index + 1
    print(ope_num)

# ABCの数
# ABCの前にあるAの分だけ追加で操作できる
# ABCの後にある(BC)の分だけ追加で操作できる

# Aがあれば覚えておく
# ABCと来たら、それより前側に隣接しているAの回数だけ追加で操作できる
# 見つかったABCをBCAに置き換える
# BCAのAから走査を継続する


Fixed submission:
def li():
    return list(map(int, input().split()))


if __name__ == "__main__":
    s = input()
    ope_num = 0
    index = 0
    a_stock = 0
    b_stock = 0
    s = list(s)
    while index < len(s):
        # print(s)
        # print(index)
        # print(ope_num)
        # print()
        si = s[index]
        if si == "A":
            if b_stock > 0:
                a_stock = 1
            else:
                a_stock = a_stock + 1
            b_stock = 0
        elif a_stock > 0:
            if b_stock == 0:
                if si == "B":
                    b_stock = 1
                else:
                    a_stock = 0
                    b_stock = 0
            elif si == "C":
                ope_num = ope_num + a_stock
                # s[index - 2 - a_stock + 1] = "B"
                # s[index - 1 - a_stock + 1] = "C"
                # s[index] = "A"
                # index = index - 1
                a_stock = a_stock
                b_stock = 0
            else:
                a_stock = 0
                b_stock = 0
        else:
            a_stock = 0
            b_stock = 0
        index = index + 1
    print(ope_num)

# ABCの数
# ABCの前にあるAの分だけ追加で操作できる
# ABCの後にある(BC)の分だけ追加で操作できる

# Aがあれば覚えておく
# ABCと来たら、それより前側に隣接しているAの回数だけ追加で操作できる
# 見つかったABCをBCAに置き換える
# BCAのAから走査を継続する

Buggy python submission (verdict: Time Limit Exceeded):
def li():
    return list(map(int, input().split()))


if __name__ == "__main__":
    s = input()
    ope_num = 0
    index = 0
    a_stock = []
    b_stock = []
    s = list(s)
    while index < len(s):
        si = s[index]
        # print(index)
        # print(si)
        # print(a_stock)
        # print(b_stock)
        # print(ope_num)
        # print()
        if si == "A":
            if len(b_stock) > 0:
                b_stock = []
                a_stock = [si]
            else:
                a_stock.append(si)
        elif si == "B" and len(a_stock) > 0 and len(b_stock) == 0:
            b_stock.append(si)
        elif si == "C" and len(a_stock) > 0 and len(b_stock) > 0:
            ope_num = ope_num + 1
            s[index - 2] = "B"
            s[index - 1] = "C"
            s[index] = "A"
            a_stock = []
            b_stock = []
            if index - 4 >= -1:
                index = index - 4
            else:
                index = index - 1
        else:
            a_stock = []
            b_stock = []
        index = index + 1
    print(ope_num)

# ABCの数
# ABCの前にあるAの分だけ追加で操作できる
# ABCの後にある(BC)の分だけ追加で操作できる

# Aがあれば覚えておく
# ABCと来たら、それより前側に隣接しているAの回数だけ追加で操作できる
# 見つかったABCをBCAに置き換える
# BCAのAから走査を継続する


Fixed submission:
def li():
    return list(map(int, input().split()))


if __name__ == "__main__":
    s = input()
    ope_num = 0
    index = 0
    a_stock = 0
    b_stock = 0
    s = list(s)
    while index < len(s):
        # print(s)
        # print(index)
        # print(ope_num)
        # print()
        si = s[index]
        if si == "A":
            if b_stock > 0:
                a_stock = 1
            else:
                a_stock = a_stock + 1
            b_stock = 0
        elif a_stock > 0:
            if b_stock == 0:
                if si == "B":
                    b_stock = 1
                else:
                    a_stock = 0
                    b_stock = 0
            elif si == "C":
                ope_num = ope_num + a_stock
                # s[index - 2 - a_stock + 1] = "B"
                # s[index - 1 - a_stock + 1] = "C"
                # s[index] = "A"
                # index = index - 1
                a_stock = a_stock
                b_stock = 0
            else:
                a_stock = 0
                b_stock = 0
        else:
            a_stock = 0
            b_stock = 0
        index = index + 1
    print(ope_num)

# ABCの数
# ABCの前にあるAの分だけ追加で操作できる
# ABCの後にある(BC)の分だけ追加で操作できる

# Aがあれば覚えておく
# ABCと来たら、それより前側に隣接しているAの回数だけ追加で操作できる
# 見つかったABCをBCAに置き換える
# BCAのAから走査を継続する