A string is finite sequence of characters over a non-empty finite set Σ.

In this problem, Σ is the set of lowercase letters.

Substring, also called factor, is a consecutive sequence of characters occurrences at least once in a string.

Now your task is simple, for two given strings, find the length of the longest common substring of them.

Here common substring means a substring of two or more strings.

Input

The input contains exactly two lines, each line consists of no more than 250000 lowercase letters, representing a string.

Output

The length of the longest common substring. If such string doesn't exist, print "0" instead.

Example

Input:alsdfkjfjkdsalfdjskalajfkdslaOutput:3

Notice: new testcases added

 

为什么都说这是后缀自动机的裸题啊。难道就我看不出来么qwq、、

正解其实比较好想，先把第一个串扔到SAM里面

对于第二个串依次枚举，如果能匹配就匹配，否则就沿着parent边走(怎么这么像AC自动机？？)，顺便记录一下最长长度

 

 

#include
     
      #include
      
       #include
       
        using namespace std;const int MAXN = 2 * 300000 + 1;char s1[MAXN], s2[MAXN];int N1, N2, tot = 1, root = 1, last = 1, len[MAXN], ch[MAXN][27], fa[MAXN];void insert(int x) {    int now = ++tot, pre = last; last = now; len[now] = len[pre] + 1;    for(; pre && !ch[pre][x]; pre = fa[pre]) ch[pre][x] = now;    if(!pre) fa[now] = root;    else {        int q = ch[pre][x];        if(len[q] == len[pre] + 1) fa[now] = q;        else {            int nows = ++tot;             memcpy(ch[nows], ch[q], sizeof(ch[q]));            len[nows] = len[pre] + 1;            fa[nows] = fa[q]; fa[now] = fa[q] = nows;            for(; pre && ch[pre][x] == q; pre = fa[pre]) ch[pre][x] = nows;                        }    } }int main() {#ifdef WIN32    freopen("a.in", "r", stdin);#endif    scanf("%s %s", s1 + 1, s2 + 1);    N1 = strlen(s1 + 1);     N2 = strlen(s2 + 1);    for(int i = 1; i <= N1; i++)         insert(s1[i] - 'a');    int ans = 0, nowlen = 0, cur = root;    for(int i = 1; i <= N2; i++, ans = max(ans, nowlen)) {        int p = s2[i] - 'a';        if(ch[cur][p]) {nowlen++, cur = ch[cur][p]; continue;}        for(; cur && !ch[cur][p]; cur = fa[cur]);        nowlen = len[cur] + 1, cur = ch[cur][p];    }    printf("%d\n", ans);    return 0;}