Koa the Koala has a binary string s of length n. Koa can perform no more than n-1 (possibly zero) operations of the following form:

In one operation Koa selects positions i and i+1 for some i with 1 ≤ i < |s| and sets s_i to max(s_i, s_{i+1}). Then Koa deletes position i+1 from s (after the removal, the remaining parts are concatenated).

Note that after every operation the length of s decreases by 1.

How many different binary strings can Koa obtain by doing no more than n-1 (possibly zero) operations modulo 10^9+7 (1000000007)?

Input

The only line of input contains binary string s (1 ≤ |s| ≤ 10^6). For all i (1 ≤ i ≤ |s|) s_i = 0 or s_i = 1.

Output

On a single line print the answer to the problem modulo 10^9+7 (1000000007).

Examples

Input


000


Output


3


Input


0101


Output


6


Input


0001111


Output


16


Input


00101100011100


Output


477

Note

In the first sample Koa can obtain binary strings: 0, 00 and 000.

In the second sample Koa can obtain binary strings: 1, 01, 11, 011, 101 and 0101. For example:

  * to obtain 01 from 0101 Koa can operate as follows: 0101 → 0(10)1 → 011 → 0(11) → 01. 
  * to obtain 11 from 0101 Koa can operate as follows: 0101 → (01)01 → 101 → 1(01) → 11. 



Parentheses denote the two positions Koa selected in each operation.
#include <bits/stdc++.h>
using namespace std;
const int p = 1000000007;
int N, cntsuf0;
char s[1000005];
int A[1000005];
int stk[1000005], len, ans[1000005];
int main() {
  scanf("%s", s + 1);
  N = strlen(s + 1);
  while (s[N] == '0') cntsuf0++, N--;
  if (!N) {
    printf("%d\n", cntsuf0);
    return 0;
  }
  int ANS = 0, SUM = 0;
  for (int i = N; i; i--) {
    int j = i;
    while (s[j - 1] == '0') j--;
    A[i] = i - j, ans[i] = SUM + 1;
    while (len && A[stk[len]] < A[i])
      SUM = (SUM + 1LL * ans[stk[len]] * (p - A[stk[len]] - 1) % p +
             1LL * ans[stk[len - 1]] * (A[stk[len]] + 1) % p) %
            p,
      len--;
    SUM = (SUM + 1LL * ans[i] * (A[i] + 1) % p +
           1LL * ans[stk[len]] * (p - A[i] - 1) % p) %
          p;
    stk[++len] = i;
    if (j == 1) ANS = 1LL * ans[i] * (A[i] + 1) % p;
    i = j;
  }
  printf("%d\n", 1LL * ANS * (cntsuf0 + 1) % p);
  return 0;
}
