Codeforces 17E Palisection 【Manacher】

Codeforces 17E Palisection

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E. Palisection

In an English class Nick had nothing to do at all, and remembered about wonderful strings called palindromes. We should remind you that a string is called a palindrome if it can be read the same way both from left to right and from right to left. Here are examples of such strings: «eye», «pop», «level», «aba», «deed», «racecar», «rotor», «madam».

Nick started to look carefully for all palindromes in the text that they were reading in the class. For each occurrence of each palindrome in the text he wrote a pair — the position of the beginning and the position of the ending of this occurrence in the text. Nick called each occurrence of each palindrome he found in the text subpalindrome. When he found all the subpalindromes, he decided to find out how many different pairs among these subpalindromes cross. Two subpalindromes cross if they cover common positions in the text. No palindrome can cross itself.

Let’s look at the actions, performed by Nick, by the example of text «babb». At first he wrote out all subpalindromes:

• «b» — 1..1

• «bab» — 1..3

• «a» — 2..2

• «b» — 3..3

• «bb» — 3..4

• «b» — 4..4

Then Nick counted the amount of different pairs among these subpalindromes that cross. These pairs were six:

1. 1..1 cross with 1..3
2. 1..3 cross with 2..2
3. 1..3 cross with 3..3
4. 1..3 cross with 3..4
5. 3..3 cross with 3..4
6. 3..4 cross with 4..4
   
   Since it’s very exhausting to perform all the described actions manually, Nick asked you to help him and write a program that can find out the amount of different subpalindrome pairs that cross. Two subpalindrome pairs are regarded as different if one of the pairs contains a subpalindrome that the other does not.

Input

The first input line contains integer n (1 ≤ n ≤ 2·106) — length of the text. The following line contains n lower-case Latin letters (from a to z).

Output

In the only line output the amount of different pairs of two subpalindromes that cross each other. Output the answer modulo 51123987.

Examples

input

4

babb

output

6

input

2

aa

output

2

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恶心死了

对于两个回文串的位置关系

我们只减少回文串右端点严格小于另一个回文串左端点的情况

所以两个串的关系最多只会被减少一次

需要排除没有意义的情况

还需要差分计算贡献，注意倒着跑

还需要单独考虑回文串和单个字符的交，还是找找规律求一个前缀和就好

前缀和：因为我们对于一个点，只求出了最长回文串，又因为这个回文串包含的小回文串（同一中心）的左右节点变化是每次加或碱2，所以可以前缀和求出

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#include<bits/stdc++.h>
using namespace std;
#define N 2000010
#define Mod 51123987
#define LL long long
int len,n;
int r[N<<1],p[N<<1],psum[N<<1];
char t[N],s[N<<1];
void Manacher(){
    int id=0,pos=0,x;
    for(int i=1;i<n;i++){
        if(pos>i)x=min(p[id*2-i],pos-i+1);
        else x=1;
        while(s[i-x]==s[i+x])x++;
        x--;
        if(x+i>pos)pos=x+i,id=i;
        p[i]=x;
    }
}
int main(){
    scanf("%d%s",&len,t);
    s[n=0]='!';
    for(int i=0;i<len;i++)s[++n]='#',s[++n]=t[i];
    s[++n]='#';s[++n]='?';
    Manacher();
    int ans=0,tmp=0;
    for(int i=1;i<n;i++){
        if(p[i]==0&&s[i]=='#')continue;
        if(p[i]==1&&s[i]!='#')continue;
        //枚举左右边界，对当前回文串包含的所有回文串进行累加
        r[i-p[i]-1]--;r[i-1]++;//差分求贡献
        tmp=(tmp+p[i]/2)%Mod;
    }
    //r差分累加
    //r第一次累加 前缀和
    //r第二次累加 计算答案
    for(int i=n-1;i>0;i--)r[i]+=r[i+1];
    for(int i=n-1;i>=1;i--){
        if(i&1)r[i]=r[i+1];
        else r[i]=(r[i]+r[i+1])%Mod;
    }
    for(int i=n-3;i>0;i--)r[i]=(r[i]+r[i+2])%Mod;
    ans=1ll*tmp*(tmp-1)/2%Mod;
    for(int i=1;i<n;i++){
        if(p[i]==0&&s[i]=='#')continue;
        if(p[i]==1&&s[i]!='#')continue;
        ans-=r[i+3]-r[i+p[i]+3];
        ans=(ans%Mod+Mod)%Mod;
    }
    //讨论回文串和单个字符的交
    for(int i=2;i<n;i++)psum[i]=(i+psum[i-2])%Mod;
    for(int i=1;i<n;i++)ans=(ans+psum[p[i]])%Mod;
    printf("%d",ans);
    return 0;
}