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;
vector<pair<int, int> > AdjList[2005];
int colour[2005];
const long long mod = 998244353;
long long N, tot, total;
int check = 1;
string s;
void dfs(int x) {
  for (auto it : AdjList[x]) {
    int exp_colour = (it.second == 0 ? colour[x] : 1 - colour[x]);
    if (colour[it.first] == -1) {
      colour[it.first] = exp_colour;
      dfs(it.first);
    } else if (colour[it.first] != exp_colour) {
      check = 0;
    }
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cin >> s;
  N = s.size();
  reverse(s.begin(), s.end());
  for (int i = 1; i <= N; ++i) {
    check = 1;
    for (int j = 0; j < 2005; ++j) {
      AdjList[j].clear();
      colour[j] = -1;
    }
    colour[N] = 1;
    for (int j = 1; j <= N; ++j) {
      AdjList[j].push_back(pair<int, int>(N - j + 1, 0));
      AdjList[N - j + 1].push_back(pair<int, int>(j, 0));
      if (s[j - 1] == '0') {
        AdjList[j].push_back(pair<int, int>(j + N, 0));
        AdjList[j + N].push_back(pair<int, int>(j, 0));
      } else if (s[j - 1] == '1') {
        AdjList[j].push_back(pair<int, int>(j + N, 1));
        AdjList[j + N].push_back(pair<int, int>(j, 1));
      }
    }
    for (int j = 1; j <= i; ++j) {
      AdjList[N + j].push_back(pair<int, int>(N + i - j + 1, 0));
      AdjList[N + i - j + 1].push_back(pair<int, int>(N + j, 0));
    }
    colour[N + i] = 1;
    for (int j = i + 1; j <= N; ++j) {
      colour[N + j] = 0;
    }
    dfs(N);
    dfs(N + i);
    for (int j = i + 1; j <= N; ++j) {
      dfs(N + j);
    }
    tot = 1;
    for (int j = 1; j <= 2 * N; ++j) {
      if (colour[j] == -1) {
        tot *= 2;
        colour[j] = 0;
        dfs(j);
        tot %= mod;
      }
    }
    if (!check) tot = 0;
    total += tot;
    total %= mod;
  }
  cout << total << '\n';
}