Problem: You are given a string s consisting of characters "1", "0", and "?". The first character of s is guaranteed to be "1". Let m be the number of characters in s.

Count the number of ways we can choose a pair of integers a, b that satisfies the following: 

  * 1 ≤ a < b < 2^m 
  * When written without leading zeros, the base-2 representations of a and b are both palindromes. 
  * The base-2 representation of bitwise XOR of a and b matches the pattern s. We say that t matches s if the lengths of t and s are the same and for every i, the i-th character of t is equal to the i-th character of s, or the i-th character of s is "?". 



Compute this count modulo 998244353. 

Input

The first line contains a single string s (1 ≤ |s| ≤ 1 000). s consists only of characters "1", "0" and "?". It is guaranteed that the first character of s is a "1".

Output

Print a single integer, the count of pairs that satisfy the conditions modulo 998244353.

Examples

Input


10110


Output


3


Input


1?0???10


Output


44


Input


1?????????????????????????????????????


Output


519569202


Input


1


Output


0

Note

For the first example, the pairs in base-2 are (111, 10001), (11, 10101), (1001, 11111).

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
struct bipartite_union_find {
  vector<int> parent;
  vector<int> size;
  vector<bool> bipartite;
  vector<bool> edge_parity;
  int components;
  bipartite_union_find(int n = 0) {
    if (n > 0) init(n);
  }
  void init(int n) {
    parent.resize(n + 1);
    size.assign(n + 1, 1);
    bipartite.assign(n + 1, true);
    edge_parity.assign(n + 1, false);
    components = n;
    for (int i = 0; i <= n; i++) parent[i] = i;
  }
  int find(int x) {
    if (x == parent[x]) return x;
    int root = find(parent[x]);
    edge_parity[x] = edge_parity[x] ^ edge_parity[parent[x]];
    return parent[x] = root;
  }
  pair<bool, bool> unite(int x, int y) {
    int x_root = find(x);
    int y_root = find(y);
    if (x_root == y_root) {
      bool consistent = edge_parity[x] != edge_parity[y];
      if (!consistent) bipartite[x_root] = false;
      return {false, consistent};
    }
    bool needed_parity = edge_parity[x] == edge_parity[y];
    x = x_root;
    y = y_root;
    if (size[x] < size[y]) swap(x, y);
    parent[y] = x;
    size[x] += size[y];
    bipartite[x] = bipartite[x] && bipartite[y];
    edge_parity[y] = needed_parity;
    components--;
    return {true, true};
  }
};
const int V_MAX = 5005;
const int MOD = 998244353;
int N, V;
string S;
bipartite_union_find UF;
bool different(int a, int b) { return UF.unite(a, b).second; }
bool same(int a, int b) {
  V++;
  return different(a, V - 1) && different(b, V - 1);
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  cin >> S;
  N = S.size();
  if (S.back() == '1') {
    cout << 0 << '\n';
    return 0;
  }
  long long total = 0;
  for (int length = 1; length < N; length++) {
    UF.init(V_MAX);
    V = N + length + 2;
    int zero = V - 2, one = V - 1;
    bool valid = different(zero, one);
    valid &= same(0, one);
    valid &= same(N - 1, one);
    valid &= same(N, one);
    valid &= same(N + length - 1, one);
    for (int i = 1; i < N - length; i++)
      if (S[i] != '?') valid &= same(i, S[i] == '0' ? zero : one);
    for (int i = N - length; i < N; i++)
      if (S[i] == '0')
        valid &= same(i, i + length);
      else if (S[i] == '1')
        valid &= different(i, i + length);
    for (int i = 0, j = N - 1; i < j; i++, j--) valid &= same(i, j);
    for (int i = 0, j = length - 1; i < j; i++, j--)
      valid &= same(N + i, N + j);
    if (valid) {
      int components = UF.components - (V_MAX - V) - 1;
      long long now = 1;
      for (int i = 0; i < components; i++) now = 2 * now % MOD;
      total = (total + now) % MOD;
    }
  }
  cout << total << '\n';
}