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 N = 2005;
int n;
int fa[N], v[N], st, ed;
char s[N];
int ans = 0;
long long qp(int x, int l) {
  int ret = 1;
  while (l) {
    if (l & 1) ret = 1ll * ret * x % mod;
    l >>= 1;
    x = 1ll * x * x % mod;
  }
  return ret;
}
int finds(int x) {
  if (fa[x] == x) return x;
  int now = finds(fa[x]);
  v[x] ^= v[fa[x]];
  fa[x] = now;
  return fa[x];
}
int merge(int x, int y, int z) {
  int fx = finds(x), fy = finds(y);
  if (fx == fy) return v[x] ^ v[y] ^ z ^ 1;
  fa[fx] = fy;
  v[fx] = v[x] ^ v[y] ^ z;
  return 1;
}
int calc(int len) {
  for (int i = 1; i <= ed; ++i) fa[i] = i, v[i] = 0;
  for (int i = 1; i <= n / 2; ++i) merge(i, n - i + 1, 0);
  for (int i = 1; i <= len / 2; ++i) merge(n + i, n + len - i + 1, 0);
  for (int i = len + 1; i <= n; ++i) merge(n + i, ed, 0);
  merge(1, st, 0);
  merge(n + 1, st, 0);
  for (int i = 1; i <= n; ++i) {
    if (s[n - i + 1] != '?') {
      if (!merge(n + i, i, s[n - i + 1] - '0')) return 0;
    }
  }
  int cnt = 0;
  for (int i = 1; i <= ed; ++i) cnt += (fa[i] == i);
  return qp(2, cnt - 1);
}
int main() {
  scanf("%s", s + 1);
  n = strlen(s + 1);
  st = 2 * n + 1;
  ed = st + 1;
  for (int i = 1; i < n; ++i) ans = (ans + calc(i)) % mod;
  printf("%d\n", ans);
  return 0;
}