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;
int i, j, k, m, n, p, u, x, a[1005], v[1005], w[1005];
string s;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> s;
  n = s.size();
  if (s[n - 1] == 49) {
    cout << 0;
    return 0;
  }
  for (s[n - 1] = 0; i < n; i++) s[i] &= 3;
  for (; ++k < n;) {
    for (i = 0; i < n; i++) v[i] = w[i] = 1;
    for (i = 0, p = 1; i < n; i++) {
      if (v[i]) {
        for (m = i, u = 0;;) {
          v[m] = 0;
          if (m == n - m - 1) {
            if (u < 0 || s[m] & 2) p = 2 * p % 998244353;
            break;
          }
          if (s[m] & 2)
            u = -1;
          else if (~u)
            u ^= s[m];
          m = n - m - 1;
          v[m] = 0;
          if (s[m] & 2 && u < 0) p = 2 * p % 998244353;
          if (m < n - k) {
            if (s[m] + u == 1) p = 0;
            break;
          }
          w[m] = 0;
          if (m == n + n - m - k - 1) {
            if (u < 0 || s[m] & 2) p = 2 * p % 998244353;
            break;
          }
          if (s[m] & 2)
            u = -1;
          else if (~u)
            u ^= s[m];
          m = n + n - m - k - 1;
          w[m] = 0;
          if (s[m] & 2 && u < 0) p = 2 * p % 998244353;
        }
      }
    }
    x = (x + p) % 998244353;
  }
  cout << x << '\n';
}