A number is skr, if and only if it's unchanged after being reversed. For example, "12321", "11" and "1" are skr numbers, but "123", "221" are not. FYW has a string of numbers, each substring can present a number, he wants to know the sum of distinct skr number in the string. FYW are not good at math, so he asks you for help.

Input

The only line contains the string of numbers SSS.

It is guaranteed that 1≤S[i]≤91 \le S[i] \le 91≤S[i]≤9, the length of SSS is less than 200000020000002000000.

Output

Print the answer modulo 100000000710000000071000000007.

樣例輸入1

111111

樣例輸出1

123456

樣例輸入2

1121

樣例輸出2

135

#include<bits/stdc++.h>
using namespace std;

typedef unsigned long long ULL;
typedef long long LL;
#define rep(i,a,b) for(int i=a;i<b;++i)
#define per(i,a,b) for(int i=b-1;i>=a;--i)

const int N=2e6+10;
const int mod=1e9+7;


char s[N];

ULL base[N*2],has[N*2];//auto
LL base10[N*2],sum[N*2];


const ULL P=1000007;
int fir[N<<1],nxt[N<<1],cnt=0;
ULL has_link[N<<1];
int HAS(ULL x){
    ULL t=x%P;
    for(int i=fir[t];i;i=nxt[i]){
        if(has_link[i]==x)return 1;
    }
    nxt[++cnt]=fir[t];
    fir[t]=cnt;
    has_link[cnt]=x;
    return 0;
}

LL solve(int l,int r){
    ULL v=has[r]-has[l-1]*base[r-l+1];
    if(HAS(v))return 0;
    LL res=(sum[r]-sum[l-1]*base10[(r-l+2)/2]%mod+mod)%mod;
    return res;
}

char Ma[N*2];
int Mp[N*2];//record radius
void Manacher(char s[],int len){
    int l=0;
    Ma[l++]='$';//head
    Ma[l++]='#';
    rep(i,0,len){
        Ma[l++]=s[i];
        Ma[l++]='#';
    }
    Ma[l]=0;//tail

    base[0]=base10[0]=1;
    sum[0]=has[0]=0;
    rep(i,1,l+1){
        base[i]=base[i-1]*P,has[i]=has[i-1]*P+Ma[i];

        base10[i]=base10[i-1]*10%mod;
        if(Ma[i]>='0'&&Ma[i]<='9') sum[i]=(sum[i-1]*10+Ma[i]-'0')%mod;
        else sum[i]=sum[i-1];
    }

    int mx=0,id=0;//record max_pos
    LL ans=0;
    rep(i,1,l){
        if(Ma[i]!='#')ans=(ans+solve(i,i))%mod;
        Mp[i]=mx>i?min(Mp[2*id-i],mx-i):1;
        while(Ma[i+Mp[i]]==Ma[i-Mp[i]]){
            if(Ma[i+Mp[i]]!='#')ans=(ans+solve(i-Mp[i],i+Mp[i]))%mod;
            ++Mp[i];//exceed mx
        }
        if(i+Mp[i]>mx){
            mx=i+Mp[i];
            id=i;
        }
    }
    printf("%lld\n",ans);
}

int main(){
    scanf("%s",s);
    int len=strlen(s);
    Manacher(s,len);
    return 0;
}