Police headquarter is monitoring signal on different frequency levels. They have got two suspiciously encoded strings s1 and s2 from two different frequencies as signals. They are suspecting that these two strings are from two different criminals and they are planning to do some evil task.

Now they are trying to find a common substring of minimum length between these two strings. The substring must occur only once in the first string, and also it must occur only once in the second string.

Given two strings s1 and s2 consist of lowercase Latin letters, find the smallest (by length) common substring p of both s1 and s2, where p is a unique substring in s1 and also in s2. See notes for formal definition of substring and uniqueness.

Input
The first line of input contains s1 and the second line contains s2 (1 ≤ |s1|, |s2| ≤ 5000). Both strings consist of lowercase Latin letters.

Output
Print the length of the smallest common unique substring of s1 and s2. If there are no common unique substrings of s1 and s2 print -1.
题目大意：
就是对于两个字符串s1和s2, 求出他们最短的一个公共字串t,满足t在s1中只出现1次, t在s2中也只出现一次, 输出t最小的长度, 如果这样的t不存在, 输出-1

思路：将这两个字符串拼成一个串，中间用一个没出现过的字符隔开（我用的空格），对合成的串求后缀数组，这样两个数组的公共子串所在的后缀在sa中应该是相邻的。通过观察，只有当这两个sa中相邻的后缀sa[i]和sa[i-1]左右的后缀都不含这个子串时，才能保证子串在两个串都只出现一次。

具体来说就是
（1）后缀sa[i]和sa[i-1]分别属于两个字符串
（2）height[i-1] < height[i] && height[i+1] < height[i]
（3）这个最小公共子串长度就是max(height[i-1], height[i+1]) + 1

整体扫过一遍height数组统计最小值即可，没有满足的情况时输出-1

#include <iostream>
#include <stdio.h>
#include <string.h>
using namespace std;
const int maxn = 2e6 + 5;
char s1[maxn], s2[maxn], s[maxn];
int n, p1, p2, m, p;
int tx[maxn], sa[maxn], rak[maxn];
int height[maxn], num1[maxn];
const int inf = 1e9;
struct node {
	int x, y, id;
}a[maxn], b[maxn];

void rsort() {
	for (int i = 1; i <= m; i++) {
		tx[i] = 0;
	}
	for (int i = 1; i <= n; i++) {
		tx[a[i].y]++;
	}
	for (int i = 1; i <= m; i++) {
		tx[i] += tx[i - 1];
	}
	for (int i = 1; i <= n; i++) {
		b[tx[a[i].y]--] = a[i];
	}
	for (int i = 1; i <= m; i++) {
		tx[i] = 0;
	}
	for (int i = 1; i <= n; i++) {
		tx[b[i].x]++;
	}
	for (int i = 1; i <= m; i++) {
		tx[i] += tx[i - 1];
	}
	for (int i = n; i >= 1; i--) {
		a[tx[b[i].x]--] = b[i];
	}
}

void solve() {
	rsort();
	p = 0;
	for (int i = 1; i <= n; i++) {
		if (a[i].x == a[i - 1].x && a[i].y == a[i - 1].y) {
			rak[a[i].id] = p;
		} else {
			rak[a[i].id] = ++p;
		}
	}
	for (int i = 1; i <= n; i++) {
		a[i].x = rak[i];
		a[i].id = sa[rak[i]] = i;
		a[i].y = 0;
	}
	m = p;
}
void ssort() {
	m = 127;
	for (int i = 1; i <= n; i++) {
		a[i].x = a[i].y = s[i];
		a[i].id = i;
	}
	solve();
	for (int j = 1; j <= n; j <<= 1) {
		for (int i = 1; i + j <= n; i++) {
			a[i].y = a[i + j].x;
		}
		solve();
		if (p == n) {
			break;
		}
	}
}	

void get_Height() {
	int k = 0;
	for (int i = 1; i <= n; i++) {
		if (k) {
			k--;
		}
		int j = sa[rak[i] - 1];
		while (s[i + k] == s[j + k]) {
			k++;
		}
		height[rak[i]] = k;
	}
}

int main() {
	scanf("%s", s1 + 1);
	scanf("%s", s2 + 1);
	p1 = strlen(s1 + 1);
	p2 = strlen(s2 + 1);
	for (int i = 1; i <= p1; i++) {
		s[++n] = s1[i];
		num1[n] = 2;
	}
	s[++n] = '#';
	for (int i = 1; i <= p2; i++) {
		s[++n] = s2[i];
		num1[n] = 1;
	}
	ssort();
	get_Height();
	int ans = inf;
	for (int i = 2; i <= n; i++) {
		
			if (num1[sa[i]] + num1[sa[i - 1]] == 3 && height[i - 1] < height[i] && height[i] > height[i + 1]) {
				ans = min(ans, max(height[i - 1], height[i + 1]) + 1);
			
		}
	}
	printf("%d\n", ans == inf ? -1 : ans);
	return 0;
}