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>
char s[1048576];
int a[1048576];
std::pair<int, int> st[1048576];
int main() {
  int n = 0;
  int base = 0;
  scanf("%s", s);
  {
    int i = 0;
    while (s[i] == '0') ++base, ++i;
    if (s[i] == 0) return printf("%d\n", base), 0;
    assert(s[i] == '1');
    ++i;
    int cnt = 0;
    while (s[i]) {
      assert(s[i] == '0' || s[i] == '1');
      ++cnt;
      if (s[i] == '1') {
        a[n++] = cnt;
        cnt = 0;
      }
      ++i;
    }
    base = (long long int)(base + 1) * (cnt + 1) % ((int)1e9 + 7);
  }
  int ns = 0;
  st[ns++] = {0, 1};
  for (int i = 0; i < n; ++i) {
    int cz = 0, cnt = 0;
    while (ns && st[ns - 1].first <= a[i]) {
      int k, c;
      std::tie(k, c) = st[--ns];
      cz = (cz + (long long int)(a[i] - k) * c) % ((int)1e9 + 7);
      cnt = (cnt + c) % ((int)1e9 + 7);
    }
    st[ns++] = {a[i], cnt};
    st[ns++] = {0, cz};
    0;
  }
  int ans = 0;
  while (ns) ans = (ans + st[--ns].second) % ((int)1e9 + 7);
  ans = (long long int)ans * base % ((int)1e9 + 7);
  printf("%d\n", ans);
  return 0;
}
