#include <bits/stdc++.h>

#include <atcoder/fenwicktree>
#include <atcoder/modint>
using namespace std;
using namespace atcoder;

#pragma GCC target("avx2")
#pragma GCC optimize("O3")
#pragma GCC optimize("unroll-loops")

using M1 = modint1000000007;
using M2 = modint998244353;
using hash_t = pair<M1, M2>;
namespace atcoder {
bool operator<(const M1 &t1, const M1 &t2) { return t1.val() < t2.val(); }
bool operator<(const M2 &t1, const M2 &t2) { return t1.val() < t2.val(); }
}  // namespace atcoder

struct Hash {
  int n;
  string s;
  long long B;
  bool calc_rev;
  fenwick_tree<M1> bit1, bit1_rev;
  fenwick_tree<M2> bit2, bit2_rev;
  vector<M1> Bp1, Bp1_inv;
  vector<M2> Bp2, Bp2_inv;

  Hash() {}
  Hash(string &s, bool calc_rev = true)
      : s(s), n(s.length()), calc_rev(calc_rev), Bp1(n + 1), Bp2(n + 1) {
    Bp1_inv.resize(n + 1), Bp2_inv.resize(n + 1);
    random_device seed;
    mt19937 engine(seed());
    B = (long long)(engine()) + 1000;
    bit1 = fenwick_tree<M1>(n);
    bit2 = fenwick_tree<M2>(n);
    M1 p1 = 1;
    M2 p2 = 1;
    Bp1[0] = Bp1_inv[0] = 1, Bp2[0] = Bp2_inv[0] = 1;
    for (int i = 0; i < n; i++) {
      bit1.add(i, p1 * s[i]);
      bit2.add(i, p2 * s[i]);
      p1 *= B, p2 *= B;
      Bp1[i + 1] = p1;
      Bp2[i + 1] = p2;
      Bp1_inv[i + 1] = p1.inv();
      Bp2_inv[i + 1] = p2.inv();
    }
    if (not calc_rev) return;
    bit1_rev = fenwick_tree<M1>(n);
    bit2_rev = fenwick_tree<M2>(n);
    for (int i = 0; i < n; i++) {
      bit1_rev.add(i, Bp1[n - i - 1] * s[i]);
      bit2_rev.add(i, Bp2[n - i - 1] * s[i]);
    }
  }

  void update(int i, char c) {
    bit1.add(i, Bp1[i] * c - bit1.sum(i, i + 1));
    bit2.add(i, Bp2[i] * c - bit2.sum(i, i + 1));
    if (not calc_rev) return;
    bit1_rev.add(i, Bp1[n - i - 1] * c - bit1_rev.sum(i, i + 1));
    bit2_rev.add(i, Bp2[n - i - 1] * c - bit2_rev.sum(i, i + 1));
  }

  hash_t query(int l, int r, bool rev = false) {
    assert(l <= r);
    if (rev) {
      assert(calc_rev);
      hash_t res = {bit1_rev.sum(l, r), bit2_rev.sum(l, r)};
      res.first *= Bp1_inv[n - r];
      res.second *= Bp2_inv[n - r];
      return res;
    }
    hash_t res = {bit1.sum(l, r), bit2.sum(l, r)};
    res.first *= Bp1_inv[l];
    res.second *= Bp2_inv[l];
    return res;
  }

  bool is_palindrome(int l, int r) {
    assert(calc_rev);
    return query(l, (l + r) / 2, false) == query((l + r + 1) / 2, r, true);
  }
};

int main() {
  string s;
  cin >> s;
  int n = s.length();
  Hash hash(s, false);
  using mint = M1;
  vector<mint> dp(n);
  vector<int> used(n);
  auto f = [&](auto self, int l, int r) -> mint {
    if (used[l]) return dp[l];
    mint res = 1;
    for (int l2 = l + 1, r2 = r - 1; l2 <= r2; l2++, r2--) {
      if (hash.query(l, l2) == hash.query(r2, r)) {
        res += self(self, l2, r2);
      }
    }
    used[l] = true;
    return dp[l] = res;
  };
  cout << f(f, 0, n).val() << "\n";
}