The repetition number of a string is defined as the maximum number R such that the string can be partitioned into R same consecutive substrings. For example, the repetition number of "ababab" is 3 and "ababa" is 1.

Given a string containing lowercase letters, you are to find a substring of it with maximum repetition number.

The input consists of multiple test cases. Each test case contains exactly one line, which
gives a non-empty string consisting of lowercase letters. The length of the string will not be greater than 100,000.

The last test case is followed by a line containing a ‘#‘.

For each test case, print a line containing the test case number( beginning with 1) followed by the substring of maximum repetition number. If there are multiple substrings of maximum repetition number, print the lexicographically smallest one.

ccabababc
daabbccaa
#

Case 1: ababab
Case 2: aa

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cctype>
#include <algorithm>

#define maxn 200010
#define maxlog 20

int Log[maxn];

struct Suf {
	char S[maxn];
	int n, rank[maxn], height[maxn], sa[maxn], Ws[maxn];
	int mnr[maxlog][maxn], mnh[maxlog][maxn];
	
	Suf() {}
	void init(const char* Str) {
		memset(S, 0, sizeof(S));
		strcpy(S + 1, Str);
		n = strlen(S + 1);
	}
	
	bool cmp(int* a, int p1, int p2, int l) {
		if(p1 + l > n && p2 + l > n) return a[p1] == a[p2];
		if(p1 + l > n || p2 + l > n) return 0;
		return a[p1] == a[p2] && a[p1+l] == a[p2+l];
	}
	
	void ssort() {
		int *x = rank, *y = height;
		int m = 0;
		memset(Ws, 0, sizeof(Ws));
		for(int i = 1; i <= n; i++) Ws[x[i] = S[i]]++, m = std::max(m, x[i]);
		for(int i = 1; i <= m; i++) Ws[i] += Ws[i-1];
		for(int i = n; i; i--) sa[Ws[x[i]]--] = i;
		for(int j = 1, pos = 0; pos < n; j <<= 1, m = pos) {
			pos = 0;
			for(int i = n - j + 1; i <= n; i++) y[++pos] = i;
			for(int i = 1; i <= n; i++) if(sa[i] > j) y[++pos] = sa[i] - j;
			for(int i = 1; i <= m; i++) Ws[i] = 0;
			for(int i = 1; i <= n; i++) Ws[x[i]]++;
			for(int i = 1; i <= m; i++) Ws[i] += Ws[i-1];
			for(int i = n; i; i--) sa[Ws[x[y[i]]]--] = y[i];
			std::swap(x, y); pos = 1; x[sa[1]] = 1;
			for(int i = 2; i <= n; i++) x[sa[i]] = cmp(y, sa[i], sa[i-1], j) ? pos : ++pos;
		}
		return ;
	}
	
	void calch() {
		for(int i = 1; i <= n; i++) rank[sa[i]] = i;
		for(int i = 1, j, k = 0; i <= n; height[rank[i++]] = k)
			for(k ? k-- : 0, j = sa[rank[i]-1]; S[j+k] == S[i+k]; k++);
		return ;
	}
	
	void debug() {
		for(int i = 1; i <= n; i++) printf("%d%c", sa[i], i < n ? ‘ ‘ : ‘\n‘);
		for(int i = 1; i <= n; i++) printf("%d%c", height[i], i < n ? ‘ ‘ : ‘\n‘);
		return ;
	}
	
	void rmq_init() {
		Log[1] = 0;
		for(int i = 2; i <= n; i++) Log[i] = Log[i>>1] + 1;
		for(int i = 1; i <= n; i++) mnr[0][i] = rank[i], mnh[0][i] = height[i];
		for(int j = 1; (1 << j) <= n; j++)
			for(int i = 1; i + (1 << j) - 1 <= n; i++) {
				mnr[j][i] = std::min(mnr[j-1][i], mnr[j-1][i+(1<<j-1)]);
				mnh[j][i] = std::min(mnh[j-1][i], mnh[j-1][i+(1<<j-1)]);
			}
		return ;
	}
	
	int qrnk(int l, int r) {
		int t = Log[r-l+1];
		return std::min(mnr[t][l], mnr[t][r-(1<<t)+1]);
	}
	
	int qhei(int l, int r) {
		if(l > r) std::swap(l, r); l++;
		int t = Log[r-l+1];
		return std::min(mnh[t][l], mnh[t][r-(1<<t)+1]);
	}
} sol1, sol2;

char inS[maxn];

int main() {
//	freopen("data.in", "r", stdin);
//	freopen("data.out", "w", stdout);
	int kase = 0;
	while(~scanf("%s", inS)) {
		if(inS[0] == ‘#‘) break;
		sol1.init(inS);
		for(int i = 0; i < (sol1.n >> 1); i++) std::swap(inS[i], inS[sol1.n-i-1]);
		sol2.init(inS);
		
		int n = sol1.n;
		sol1.ssort(); sol1.calch(); sol1.rmq_init();
		sol2.ssort(); sol2.calch(); sol2.rmq_init();
//		sol1.debug(); sol2.debug();
		int st = sol1.sa[sol1.qrnk(1,n)], len = 1, rpt = 1;
		for(int L = 1; L <= n; L++) {
			for(int i = L; i + L <= n; i += L) {
				if(sol1.S[i] == sol1.S[i+L]) {
					int r = sol1.qhei(sol1.rank[i], sol1.rank[i+L]) - 1,
						l = sol2.qhei(sol2.rank[n-i+1], sol2.rank[n-i-L+1]) - 1,
						tmp = r / L + 1 + l / L + 1;
					if(rpt < tmp) rpt = tmp, st = sol1.sa[sol1.qrnk(i-l,i-l+(l+r+1)%L)], len = rpt * L;
					else if(rpt == tmp) {
						int pos = sol1.sa[sol1.qrnk(i-l,i-l+(l+r+1)%L)];
						if(sol1.rank[st] > sol1.rank[pos]) st = pos, len = rpt * L;
					}
				}
			}
		}
		
		printf("Case %d: ", ++kase);
		for(int i = st; i <= st + len - 1; i++) putchar(sol1.S[i]); putchar(‘\n‘);
	}
	
	return 0;
}