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 mod = 1e9 + 7;
int main() {
  std::ios::sync_with_stdio(false);
  std::cin.tie(0);
  string s;
  cin >> s;
  int n = s.size();
  if (s == string(n, '0')) {
    cout << n << endl;
    return 0;
  }
  int s0 = 0, s1 = n;
  while (s[s0] == '0') ++s0;
  while (s[s1 - 1] == '0') --s1;
  s = s.substr(s0, s1 - s0 - 1);
  s1 = n - s1;
  n = s.size();
  vector<long long> c(n + 1, 0);
  long long sum = 0;
  for (int i = 0; i < n; ++i) {
    assert(s[i] == '1');
    int j = i + 1;
    while (j < n && s[j] == '0') ++j;
    int k = j - i;
    long long nsum = sum;
    for (int t = 0; t < k; ++t) {
      nsum = (nsum + sum + 1 - c[t]) % mod;
      c[t] = (sum + 1) % mod;
    }
    sum = nsum;
    i = j - 1;
  }
  cerr << sum << ' ' << s0 << ' ' << s1 << endl;
  cout << ((sum + 1) * (s0 + 1) % mod * (s1 + 1) % mod + mod) % mod << endl;
  return 0;
}
