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;
const int mod = 998244353;
const int mxn = 4000;
int n;
int a[mxn];
string second;
vector<int> g[mxn];
void edg(int x, int y) {
  g[x].push_back(y);
  g[y].push_back(x);
}
int dfs(int c) {
  int ret = 1;
  for (int i : g[c]) {
    if (!~a[i])
      a[i] = !a[c], ret &= dfs(i);
    else if (a[c] == a[i])
      ret = 0;
  }
  return ret;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cin >> second;
  n = second.size(), reverse(second.begin(), second.end());
  long long ret = 0;
  for (int i = 1; i < n; i++) {
    long long x = 1;
    for (int j = 0; j < 2 * (n + i); j++) a[j] = -1, g[j].clear();
    a[2 * n + i - 1] = 1;
    for (int j = 0; j < n; j++) {
      edg(j, n + j);
      edg(j, 2 * n - j - 1);
      edg(n + j, n - j - 1);
      if (j < i) {
        edg(2 * n + j, 2 * n + i + j);
        edg(2 * n + j, 2 * (n + i) - j - 1);
        edg(2 * n + i + j, 2 * n + i - j - 1);
        if (second[j] == '0') {
          edg(j, 2 * n + i + j);
          edg(n + j, 2 * n + j);
        } else if (second[j] == '1') {
          edg(j, 2 * n + j);
          edg(n + j, 2 * n + i + j);
        }
      } else if (second[j] != '?') {
        a[j] = second[j] & 1;
      }
    }
    for (int j = 0; j < 2 * (n + i); j++)
      if (~a[j]) x *= dfs(j);
    for (int j = 0; j < 2 * (n + i); j++) {
      if (!~a[j]) a[j] = 0, (x *= 2 * dfs(j)) %= mod;
    }
    (ret += x) %= mod;
  }
  cout << ret << '\n';
  return 0;
}