Problem:
Score : 1600 points

Problem Statement

Taichi thinks a binary string X of odd length N is beautiful if it is possible to apply the following operation \frac{N-1}{2} times so that the only character of the resulting string is 1 :

Choose three consecutive bits of X and replace them by their median. For example, we can turn 00110 into 010 by applying the operation to the middle three bits.

Taichi has a string S consisting of characters 0, 1 and ?. Taichi wants to know the number of ways to replace the question marks with 1 or 0 so that the resulting string is beautiful, modulo 10^{9} + 7.

Constraints

1 \leq |S| \leq 300000

|S| is odd.

All characters of S are either 0, 1 or ?.

Input

Input is given from Standard Input in the following format:

S

Output

Print the number of ways to replace the question marks so that the resulting string is beautiful, modulo 10^{9} + 7.

Sample Input 1

1??00

Sample Output 1

2

There are 4 ways to replace the question marks with 0 or 1 :

11100 : This string is beautiful because we can first perform the operation on the last 3 bits to get 110 and then on the only 3 bits to get 1.

11000 : This string is beautiful because we can first perform the operation on the last 3 bits to get 110 and then on the only 3 bits to get 1.

10100 : This string is not beautiful because there is no sequence of operations such that the final string is 1.

10000 : This string is not beautiful because there is no sequence of operations such that the final string is 1.

Thus, there are 2 ways to form a beautiful string.

Sample Input 2

?

Sample Output 2

1

In this case, 1 is the only beautiful string.

Sample Input 3

?0101???10???00?1???????????????0????????????1????0

Sample Output 3

402589311

Remember to output your answer modulo 10^{9} + 7.

Solution in C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
#define ld double
#define ull unsigned long long
#define ll long long
#define pii pair <int, int>
#define iiii pair <int, pii >
#define mp make_pair
#define INF 1000000000
#define MOD 1000000007
#define rep(i, x) for(int (i) = 0; (i) < (x); (i)++)
inline int getint() {
  int x = 0, p = 1; char c = getchar();	
  while (c <= 32) c = getchar();
  if (c == 45) p = -p, c = getchar();
  while (c > 32) x = x * 10 + c - 48, c = getchar();
  return x * p;
}
using namespace std;
//ruogu_alter
int to[8][2] = {
  1, 4, 
  2, 3,
  1, 1,
  1, 4,
  5, 6,
  7, 4,
  6, 6,
  5, 5       
};
const int N = 3e5 + 5;
char s[N];
int f[N][8];
//
int main() {
  scanf("%s", s + 1);
  int n = strlen(s + 1);
  f[0][0] = 1;
  for (int i = 1; i <= n; i++) rep(j, 8) {
    if (s[i] != '0') (f[i][to[j][1]] += f[i - 1][j]) %= MOD;
    if (s[i] != '1') (f[i][to[j][0]] += f[i - 1][j]) %= MOD;
  }
  cout << (f[n][4] + f[n][6]) % MOD;
  return 0;
}