Description

A word is called a palindrome if we read from right to left is as same as we read from left to right. For example, "dad", "eye" and "racecar" are all palindromes, but "odd", "see" and "orange" are not palindromes.

Given n strings, you can generate n × n pairs of them and concatenate the pairs into single words. The task is to count how many of the so generated words are palindromes.

Input

The first line of input file contains the number of strings n. The following n lines describe each string:

The i+1-th line contains the length of the i

You can assume that the total length of all strings will not exceed 2,000,000. Two strings in different line may be the same.

Output

Print out only one integer, the number of palindromes.

Sample Input

3
1 a
2 ab
2 ba

Sample Output

5

Hint

#include<cstdio>
#include<cstring>
#include<vector>
typedef long long LL;
using namespace std;
const int N=2000005;
struct node{
    int sum,son[26],cnt;
}no[N];
struct str{
    int start,en;
    str(int s,int e):start(s),en(e){};
    str(){};
};
vector<str>ve;
int fla[N][2],nexta[N],tot=0,n,pre,l;
char s[N],t[N];
void getNext(char *a,int la){
    int i=0,j=-1;
    nexta[0]=-1;
    while(i<la){
        if(j==-1||a[i]==a[j]) nexta[++i]=++j;
        else j=nexta[j];
    }
}
void kmp(int flag,char *a,char *b,int la,int start){
    int i=0,j=0;
    while(i<la){
        if(j==-1||a[j]==b[i]) ++i,++j;
        else j=nexta[j];
    }
    int pre=j;
    if(!flag) {
        while(pre){
            fla[start+pre-1][0]=1;
            pre=nexta[pre];
        }
    }
    else {
        while(pre){
            fla[start+l-pre][1]=1;
            pre=nexta[pre];
        }
    }
}
void insertTire(int pre,char *a,int start,int la)//儲存前綴
{
     for(int i=0;i<l;i++)
     {
         if(!no[pre].son[a[i]-‘a‘])
         {
             tot++;
             no[pre].son[a[i]-‘a‘] = tot;
             pre = tot;
         }
         else pre = no[pre].son[a[i]-‘a‘];
         if(i+1<la)no[pre].cnt+=fla[start+i+1][1];
     }
     no[pre].sum++;
}
LL findTrie(int start,int en,int pre){
    int sym=true;
    LL ans=0;
    int l=en-start;
    for(int i=start;i<en;++i) {
        if(no[pre].son[t[i]-‘a‘]) {
            pre=no[pre].son[t[i]-‘a‘];
            if(fla[start+l-(i-start+1)-1][0]||i==en-1) ans+=no[pre].sum;
        }
        else {
            sym=0;break;
        }
    }
    if(sym) ans+=no[pre].cnt;
    return ans;
}
int main(){
    pre=0;
    LL ans=0;
    for(scanf("%d",&n);n--;){
        scanf("%d %s",&l,s+pre);
        ve.push_back(str(pre,pre+l));
        for(int i=pre;i<pre+l;++i) t[i]=s[pre+l-(i-pre+1)];
        getNext(s+pre,l);
        kmp(0,s+pre,t+pre,l,pre);
        getNext(t+pre,l);
        kmp(1,t+pre,s+pre,l,pre);
        insertTire(0,s+pre,pre,l);
        pre+=l;
    }
    for(int i=0;i<(int)ve.size();++i) ans+=findTrie(ve[i].start,ve[i].en,0);
    printf("%I64d\n",ans);
}

poj3376 KMP+字典樹求回文串數量(n*n)