题意：求本质不同的回文子串的和

题解：先构造pam，然后根据pam的原理（ch表示在该节点表示的回文串两侧加上该字符）对于每个节点维护一个表示该节点字符串的值，加起来即可

//#pragma GCC optimize(2)

//#pragma GCC optimize(3)

//#pragma GCC optimize(4)

//#pragma GCC optimize("unroll-loops")

//#pragma comment(linker, "/stack:200000000")

//#pragma GCC optimize("Ofast,no-stack-protector")

//#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")

#include<bits/stdc++.h>

#define fi first

#define se second

#define db double

#define mp make_pair

#define pb push_back

#define pi acos(-1.0)

#define ll long long

#define vi vector<int>

#define mod 1000000007

#define ld long double

#define C 0.5772156649

#define ls l,m,rt<<1

#define rs m+1,r,rt<<1|1

#define pll pair<ll,ll>

#define pil pair<int,ll>

#define pli pair<ll,int>

#define pii pair<int,int>

#define LL long long

//#define cd complex<double>

#define ull unsigned long long

#define base 1000000000000000000

#define Max(a,b) ((a)>(b)?(a):(b))

#define Min(a,b) ((a)<(b)?(a):(b))

#define fin freopen("a.txt","r",stdin)

#define fout freopen("a.txt","w",stdout)

#define fio ios::sync_with_stdio(false);cin.tie(0)

template<typename T>

inline T const& MAX(T const &a,T const &b){return a>b?a:b;}

template<typename T>

inline T const& MIN(T const &a,T const &b){return a<b?a:b;}

inline void add(ll &a,ll b){a+=b;if(a>=mod)a-=mod;}

inline void sub(ll &a,ll b){a-=b;if(a<0)a+=mod;}

inline ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}

inline ll qp(ll a,ll b){ll ans=1;while(b){if(b&1)ans=ans*a%mod;a=a*a%mod,b>>=1;}return ans;}

inline ll qp(ll a,ll b,ll c){ll ans=1;while(b){if(b&1)ans=ans*a%c;a=a*a%c,b>>=1;}return ans;}

using namespace std;

const double eps=1e-8;

const ll INF=0x3f3f3f3f3f3f3f3f;

const int N=2000000+10,maxn=1000000+10,inf=0x3f3f3f3f;

struct PAM{

    int ch[N][10],fail[N],cnt[N],num[N],len[N],s[N],le[N];

    ll val[N],ans[N];

    int last,n,p;

    int newnode(int w,int c)

    {

        for(int i=0;i<10;i++)ch[p][i] = 0;

        cnt[p] = num[p] = 0;

        len[p] = w;val[p]=c;

        return p++;

    }

    void init()

    {

        p = last = n = 0;

        newnode(0,0);

        newnode(-1,0);

        s[n] = -1;

        fail[0] = 1;

    }

    int getfail(int x)

    {

        while(s[n-len[x]-1] != s[n]) x = fail[x];

        return x;

    }

    void add(int c)

    {

        s[++n] = c;

        int cur = getfail(last);

        if(!ch[cur][c]){

            int now = newnode(len[cur]+2,c);

            fail[now] = ch[getfail(fail[cur])][c];

            ch[cur][c] = now;

            num[now] = num[fail[now]] + 1;

        }

        last = ch[cur][c];

        cnt[last]++;

    }

    void cal()

    {

        ll pp=0;

        for(int i=0;i<p;i++)

        {

            for(int j=0;j<10;j++)if(ch[i][j])

            {

                if(ans[i])ans[ch[i][j]]=(ans[i]*10%mod+val[ch[i][j]]+qp(10,1+le[i])*val[ch[i][j]]%mod)%mod,

                    le[ch[i][j]]=le[i]+2;

                else

                {

                    if(len[i]&1)ans[ch[i][j]]=val[ch[i][j]],le[ch[i][j]]=1;

                    else ans[ch[i][j]]=(val[ch[i][j]]+val[ch[i][j]]*10%mod)%mod,le[ch[i][j]]=2;

                }

            }

            pp+=ans[i];if(pp>=mod)pp-=mod;

        }

        printf("%lld\n",pp);

    }

}pam;

char s[N];

int main()

{

    pam.init();

    scanf("%s",s);

    int n=strlen(s);

    for(int i=0;i<n;i++)pam.add(s[i]-'0');

    pam.cal();

    return 0;

}

/********************

********************/