大家好，欢迎来到IT知识分享网。

(CodeForces 432D)Prefixes and Suffixes

time limit per test:1 second

memory limit per test:256 megabytes

input:standard input

output:standard output

You have a string s = s1s2…s|s|, where |s| is the length of string s, and si its i-th character.

Let’s introduce several definitions:
• A substring s[i..j] (1 ≤ i ≤ j ≤ |s|) of string s is string sisi + 1…sj.
• The prefix of string s of length l (1 ≤ l ≤ |s|) is string s[1..l].
• The suffix of string s of length l (1 ≤ l ≤ |s|) is string s[|s| - l + 1..|s|].

Your task is, for any prefix of string s which matches a suffix of string s, print the number of times it occurs in string s as a substring.

Input

The single line contains a sequence of characters s1s2…s|s| (1 ≤ |s| ≤ 105) — string s. The string only consists of uppercase English letters.

Output

In the first line, print integer k (0 ≤ k ≤ |s|) — the number of prefixes that match a suffix of string s. Next print k lines, in each line print two integers li ci. Numbers li ci mean that the prefix of the length li matches the suffix of length li and occurs in string s as a substring ci times. Print pairs li ci in the order of increasing li.

Examples

Input

ABACABA

Output

3
1 4
3 2
7 1

Input

AAA

Output

3
1 3
2 2
3 1

题目大意：给定一个字符串，求它既是前缀又是后缀的子串，同时求出这个子串在字符串中出现的次数。（ps：按字符串长度递增输出）

思路：对于一个字符串的Next数组记录了它的最长前缀，现用cnt数组记录其出现的次数，那么递归式：cnt[Next[i]]+=cnt[i]就是其出现的次数。

#include<cstdio>
#include<cstring>
using namespace std;

const int maxn=100005;
int l[maxn],cnt[maxn],ans[maxn];
char s[maxn];
int Next[maxn];

void getNext(int n)
{
    Next[0]=Next[1]=0;
    for(int i=1;i<n;i++)
    {
        int j=Next[i];
        while(j&&s[i]!=s[j]) j=Next[j];
        if(s[i]==s[j]) Next[i+1]=j+1;
        else Next[i+1]=0;
    }
}

int main()
{
    while(~scanf("%s",s))
    {
        int n=strlen(s);
        getNext(n);
        for(int i=0;i<=n;i++) cnt[i]=1;
        for(int i=n;i>=1;i--) cnt[Next[i]]+=cnt[i];
        int tot=0;
        for(int i=n;i;i=Next[i])
        {
            ans[tot]=cnt[i];
            l[tot++]=i;
        }
        printf("%d\n",tot);
        for(int i=tot-1;i>=0;i--) printf("%d %d\n",l[i],ans[i]);
    }
    return 0;
}