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;
int pw(int x, int y) {
  int res = 1;
  while (y) {
    if (y & 1) res = 1ll * res * x % mod;
    x = 1ll * x * x % mod;
    y >>= 1;
  }
  return res;
}
const int N = 1000;
int n;
char s[N + 5];
int fa[4 * N + 3];
int id(int x, int y, int z) { return 2 * ((x - 1) * n + y) - z; }
int root(int p) { return fa[p] == p ? p : fa[p] = root(fa[p]); }
void merge(int p, int q) { fa[root(p)] = root(q); }
int main() {
  scanf("%s", s + 1);
  n = strlen(s + 1);
  int ans = 0;
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= 4 * n + 2; j++) fa[j] = j;
    merge(id(1, 1, 1), 4 * n + 1);
    merge(id(1, 1, 0), 4 * n + 2);
    for (int j = 1; j <= n / 2; j++)
      merge(id(1, j, 0), id(1, n + 1 - j, 0)),
          merge(id(1, j, 1), id(1, n + 1 - j, 1));
    for (int j = 1; j < i; j++)
      merge(id(2, j, 0), 4 * n + 1), merge(id(2, j, 1), 4 * n + 2);
    merge(id(2, i, 1), 4 * n + 1);
    merge(id(2, i, 0), 4 * n + 2);
    for (int j = i; j <= (i - 1 + n) / 2; j++)
      merge(id(2, j, 0), id(2, n + i - j, 0)),
          merge(id(2, j, 1), id(2, n + i - j, 1));
    for (int j = 1; j <= n; j++)
      if (s[j] == '1')
        merge(id(1, j, 0), id(2, j, 1)), merge(id(1, j, 1), id(2, j, 0));
      else if (s[j] == '0')
        merge(id(1, j, 0), id(2, j, 0)), merge(id(1, j, 1), id(2, j, 1));
    bool ok = true;
    for (int j = 1; j < 4 * n + 2; j += 2)
      if (root(j) == root(j + 1)) {
        ok = false;
        break;
      }
    if (!ok) continue;
    int cnt = 0;
    for (int j = 1; j <= 4 * n + 2; j++) cnt += fa[j] == j;
    (ans += pw(2, cnt / 2 - 1)) %= mod;
  }
  printf("%d", ans);
  return 0;
}