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])