Problem: 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.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int N = 1000005;
char S[N];
int n, a[N], o, f[N];
int main() {
  scanf("%s", S + 1), n = strlen(S + 1);
  register int i, j, x = 0, s;
  for (i = 1; i <= n; ++i) S[i] == '1' ? a[++*a] = x, x = 0 : ++x;
  if (!a[0]) return printf("%d", n), 0;
  for (i = 2, o = 1, s = 0; i <= a[0];
       o = s ^ (1000000007 - 1) ? s + 1 : 0, ++i)
    for (j = 0; j <= a[i]; ++j)
      s = s - f[j] + o,
      s = (s >= 1000000007 ? s - 1000000007 : (s < 0 ? s + 1000000007 : s)),
      f[j] = o;
  printf("%d", (1ll + a[1]) * (x + 1ll) % 1000000007 * o % 1000000007);
  return 0;
}