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 MX = 1000005, mod = 1000000007;
int n, sig[MX], f, suf;
string s;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> s;
  n = s.size();
  sig[n] = -1;
  for (int i = n - 1; i >= 0; i--) {
    sig[i] = sig[i + 1];
    if (s[i] == '1') sig[i] = i, f = 1;
  }
  if (!f) {
    cout << n << '\n';
    return 0;
  }
  for (int i = n - 1; i >= 0 && s[i] == '0'; i--, suf++)
    ;
  long long acu = 0, last = 0, cn = 0;
  stack<pair<int, int> > st;
  for (int i = n - 1 - suf; i >= sig[0]; i--) {
    (acu += last) %= mod;
    if (s[i] == '0') {
      cn++;
      if (st.size()) {
        (acu -= st.top().second) %= mod;
        if (cn == st.top().first) st.pop();
      }
    } else {
      long long t = acu;
      if (cn) st.emplace(cn, last);
      (acu -= last) %= mod;
      last = (t + suf + 1) % mod;
      cn = 0;
    }
    if (i == sig[0])
      cout << (((((i + 1) * last) % (mod)) + (mod)) % (mod)) << '\n';
  }
  return 0;
}
