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;
string s;
int n;
vector<int> a;
vector<int> f;
vector<int> g;
int main(void) {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cin >> s;
  n = s.size();
  int l = 0;
  while (l < n && s[l] == '0') ++l;
  if (l == n) return cout << n << endl, 0;
  int r = n - 1;
  while (s[r] == '0') --r;
  int mul = 1ll * (l + 1) * (n - r) % mod;
  int cur = 0;
  for (int i = l + 1; i <= r; ++i) {
    if (s[i] == '0')
      ++cur;
    else {
      a.push_back(cur);
      cur = 0;
    }
  }
  if (a.empty()) return cout << mul << endl, 0;
  int m = a.size();
  f.assign(m + 1, 0);
  g.assign(n + 1, 0);
  int sum = 1;
  f[m] = 1;
  for (int i = m - 1; i >= 0; --i) {
    for (int j = 0; j <= a[i]; ++j) {
      sum = (sum + mod - g[j]) % mod;
      g[j] = f[i + 1];
      sum = (sum + g[j]) % mod;
    }
    f[i] = sum;
  }
  cout << 1ll * f[0] * mul % mod << endl;
  return 0;
}
