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 long long INF = 1e9;
inline long long read() {
  long long X = 0, w = 0;
  char ch = 0;
  while (!isdigit(ch)) {
    w |= ch == '-';
    ch = getchar();
  }
  while (isdigit(ch)) X = (X << 3) + (X << 1) + (ch ^ 48), ch = getchar();
  return w ? -X : X;
}
long long mod = 998244353;
long long pw(long long x, long long y) {
  if (!y) return 1;
  if (y == 1) return x;
  long long mid = pw(x, y >> 1);
  if (y & 1) return 1ll * mid * mid % mod * x % mod;
  return 1ll * mid * mid % mod;
}
string s;
long long ans;
long long n;
vector<pair<long long, long long> > edg[10005];
long long col[100005];
bool vis[100005];
long long ok = 1;
void dfs(long long x, long long curcol) {
  if (vis[x]) {
    if (col[x] != curcol) {
      ok = false;
    }
    return;
  }
  vis[x] = 1;
  col[x] = curcol;
  for (long long i = 0; i < edg[x].size(); i++) {
    dfs(edg[x][i].first, curcol ^ edg[x][i].second);
  }
}
long long fuc() {
  ok = 1;
  long long con = 0;
  for (long long i = 0; i <= 2 * n + 1; i++) {
    if (!vis[i]) {
      dfs(i, 0);
      ++con;
    }
  }
  if (!ok) {
    return 0;
  }
  return pw(2, con - 1);
}
long long solve(long long m) {
  memset(vis, 0, sizeof vis);
  memset(col, -1, sizeof col);
  for (long long i = 0; i <= 2005; i++) {
    edg[i].clear();
  }
  for (long long i = 0; i < n; i++) {
    edg[i].push_back(make_pair(n - 1 - i, 0));
  }
  for (long long i = 0; i < m; i++) {
    edg[i + n].push_back(make_pair(n + (m - 1 - i), 0));
  }
  for (long long i = m; i < n; i++) {
    edg[i + n].push_back(make_pair(2 * n, 0));
    edg[2 * n].push_back(make_pair(i + n, 0));
  }
  for (long long i = 0; i < n; i++) {
    if (s[n - 1 - i] == '0') {
      edg[i].push_back(make_pair(i + n, 0));
      edg[i + n].push_back(make_pair(i, 0));
    } else if (s[n - 1 - i] == '1') {
      edg[i].push_back(make_pair(i + n, 1));
      edg[i + n].push_back(make_pair(i, 1));
    }
  }
  edg[2 * n].push_back(make_pair(2 * n + 1, 1));
  edg[2 * n + 1].push_back(make_pair(2 * n, 1));
  edg[n].push_back(make_pair(2 * n + 1, 0));
  edg[2 * n + 1].push_back(make_pair(n, 0));
  edg[0].push_back(make_pair(2 * n + 1, 0));
  edg[2 * n + 1].push_back(make_pair(0, 0));
  return fuc();
}
int main() {
  cin >> s;
  n = s.size();
  for (long long i = 1; i < n; i++) {
    ans = (ans + solve(i)) % mod;
  }
  cout << ans << endl;
  return 0;
}