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 md = 1000000007, MX = 1000100;
int n, m, i, lft, cur, sum, pos, a[MX];
long long rgh;
char s[MX];
bool was;
int main() {
  scanf("%s", s);
  n = strlen(s);
  for (lft = rgh = -1, i = 0; i < n; i++)
    if (s[i] == '1') {
      was = true;
      rgh = i;
    } else if (!was)
      lft = i;
  if (!was) {
    printf("%d\n", n);
    return 0;
  }
  a[0] = sum = 1;
  for (i = lft + 1; i <= rgh; i++)
    if (s[i] == '0') {
      ++pos;
      sum = (sum + md - a[pos]) % md;
      a[pos] = cur;
      sum = (sum + a[pos]) % md;
    } else {
      cur = (cur + sum) % md;
      pos = 0;
    }
  cur = (cur * (lft + 2LL)) % md;
  cur = (cur * (n - rgh)) % md;
  printf("%d\n", cur);
  return 0;
}
