s = input()

n = len(s)
t = ""
for ss in s:
    t += ss
    t += '0'

def manacher(S):
    # 最長回文 O(n)
    # R[i] := i 文字目を中心とする最長の回文の半径（自身を含む）
    # 偶数長の回文を検出するには "$a$b$a$a$b$" のようにダミーを挟む
    # 検証: https://atcoder.jp/contests/wupc2019/submissions/8665857
    # 左右で違う条件: https://atcoder.jp/contests/code-thanks-festival-2014-a-open/submissions/12911822
    c, r, n = 0, 0, len(S)  # center, radius, length
    R = [0]*n
    while c < n:
        while c-r >= 0 and c+r < n and S[c-r] == S[c+r]:
            r += 1
        R[c] = r
        d = 1  # distance from center
        while c-d >= 0 and c+d < n and d+R[c-d] < r:
            R[c+d] = R[c-d]
            d += 1
        c += d
        r -= d
    return R

R = manacher(t)

def is_palindrome(x, y):
    l = 2 * x
    r = 2 * y
    mid = (l + r) // 2
    return R[mid] >= mid - l + 1

DP = [0 for _ in range(n + 1)]
DP[0] = 10**18
for i in range(n):
    for j in range(i, n):
        if is_palindrome(i, j):
            DP[j + 1] = max(DP[j + 1], min(DP[i], j - i + 1))
print(DP[n])