#include<bits/stdc++.h>
using namespace std;
#define rep(i,n) for(ll i=0;i<n;i++)
#define repl(i,l,r) for(ll i=(l);i<(r);i++)
#define per(i,n) for(ll i=(n)-1;i>=0;i--)
#define perl(i,r,l) for(ll i=r-1;i>=l;i--)
#define fi first
#define se second
#define pb push_back
#define ins insert
#define pqueue(x) priority_queue<x,vector<x>,greater<x>>
#define all(x) (x).begin(),(x).end()
#define CST(x) cout<<fixed<<setprecision(x)
#define vtpl(x,y,z) vector<tuple<x,y,z>>
#define rev(x) reverse(x);
using ll=long long;
using vl=vector<ll>;
using vvl=vector<vector<ll>>;
using pl=pair<ll,ll>;
using vpl=vector<pl>;
using vvpl=vector<vpl>;
const ll MOD=1000000007;
const ll MOD9=998244353;
const int inf=1e9+10;
const ll INF=4e18;
const ll dy[9]={0,1,-1,0,1,1,-1,-1,0};
const ll dx[9]={1,0,0,-1,1,-1,1,-1,0};
template<class T> inline bool chmin(T& a, T b) {
    if (a > b) {
        a = b;
        return true;
    }
    return false;
}
template<class T> inline bool chmax(T& a, T b) {
    if (a < b) {
        a = b;
        return true;
    }
    return false;
}
const int mod = 1000000007;
const int max_n = 200005;
struct mint {
  ll x; // typedef long long ll;
  mint(ll x=0):x((x%mod+mod)%mod){}
  mint operator-() const { return mint(-x);}
  mint& operator+=(const mint a) {
    if ((x += a.x) >= mod) x -= mod;
    return *this;
  }
  mint& operator-=(const mint a) {
    if ((x += mod-a.x) >= mod) x -= mod;
    return *this;
  }
  mint& operator*=(const mint a) { (x *= a.x) %= mod; return *this;}
  mint operator+(const mint a) const { return mint(*this) += a;}
  mint operator-(const mint a) const { return mint(*this) -= a;}
  mint operator*(const mint a) const { return mint(*this) *= a;}
  mint pow(ll t) const {
    if (!t) return 1;
    mint a = pow(t>>1);
    a *= a;
    if (t&1) a *= *this;
    return a;
  }
  bool operator==(const mint &p) const { return x == p.x; }
  bool operator!=(const mint &p) const { return x != p.x; }
  // for prime mod
  mint inv() const { return pow(mod-2);}
  mint& operator/=(const mint a) { return *this *= a.inv();}
  mint operator/(const mint a) const { return mint(*this) /= a;}
};
istream& operator>>(istream& is, mint& a) { return is >> a.x;}
ostream& operator<<(ostream& os, const mint& a) { return os << a.x;}
using vm=vector<mint>;
using vvm=vector<vm>;


struct RollingHash {
    static const int base1 = 1007, base2 = 2009;
    static const int mod1 = 1000000007, mod2 = 1000000009;
    vector<long long> hash1, hash2, power1, power2;

    // construct
    RollingHash(const string &S) {
        int n = (int)S.size();
        hash1.assign(n+1, 0);
        hash2.assign(n+1, 0);
        power1.assign(n+1, 1);
        power2.assign(n+1, 1);
        for (int i = 0; i < n; ++i) {
            hash1[i+1] = (hash1[i] * base1 + S[i]) % mod1;
            hash2[i+1] = (hash2[i] * base2 + S[i]) % mod2;
            power1[i+1] = (power1[i] * base1) % mod1;
            power2[i+1] = (power2[i] * base2) % mod2;
        }
    }
    
    // get hash of S[left:right)
    inline pair<long long, long long> get(int l, int r) const {
        long long res1 = hash1[r] - hash1[l] * power1[r-l] % mod1;
        if (res1 < 0) res1 += mod1;
        long long res2 = hash2[r] - hash2[l] * power2[r-l] % mod2;
        if (res2 < 0) res2 += mod2;
        return {res1, res2};
    }
    using pl=pair<long long,long long>;
    inline pl connect(pl h1,pl h2,int h2len){
        pl ret;
        ret.first=(h1.first*power1[h2len]+h2.first)%mod1;
        ret.second=(h1.second*power2[h2len]+h2.second)%mod2;
        return ret;
    }

    // get lcp of S[a:] and T[b:]
    inline int getLCP(int a, int b) const {
        int len = min((int)hash1.size()-a, (int)hash1.size()-b);
        int low = 0, high = len;
        while (high - low > 1) {
            int mid = (low + high) >> 1;
            if (get(a, a+mid) != get(b, b+mid)) high = mid;
            else low = mid;
        }
        return low;
    }
};
ll seen[5010];
mint memo[5010];
ll n;
mint solve(ll l,RollingHash &rh){//lから始まる
    if(seen[l])return memo[l];
    mint ans=1;
    for(int i=1;;i++){
        ll right=l+i;//[l,l+i)
        ll left=n-i-l;//[n-l-i,n-l)
        if(left<right)break;
        if(rh.get(l,l+i)==rh.get(n-l-i,n-l)){
            ans+=solve(l+i,rh);
        }
    }
    seen[l]=1;memo[l]=ans;
    return ans;
}
int main(){
    string s;cin >> s;
    RollingHash rh(s);
    rep(i,5010)seen[i]=0;
    n=s.size();
    cout << solve(0,rh) << endl;
}