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;
long long ans = 0;
vector<pair<long long, long long> > stk = {{(1 << 29), 0}};
vector<int> run_length;
char s[1100000];
int main() {
  scanf("%s", s);
  int n = strlen(s);
  int zero_cnt = 0;
  for (int i = 0; i < n; i++) {
    if (s[i] == '0')
      zero_cnt++;
    else {
      run_length.push_back(zero_cnt);
      zero_cnt = 0;
    }
  }
  if (run_length.empty()) {
    cout << zero_cnt << endl;
    return 0;
  }
  for (auto l : run_length) {
    if (stk.back() == pair<long long, long long>((1 << 29), 0)) {
      stk.push_back({l, l + 1});
      continue;
    }
    long long last = -1;
    long long val = 0;
    long long merged = 0;
    while (stk.back().first < l) {
      (val += stk.back().second * (l - last) % 1000000007) %= 1000000007;
      (merged += stk.back().second) %= 1000000007;
      last = stk.back().first;
      stk.pop_back();
    }
    (val += stk.back().second * (l - last) % 1000000007) %= 1000000007;
    (stk.back().second += merged) %= 1000000007;
    stk.push_back({l, val});
  }
  for (auto x : stk) (ans += x.second) %= 1000000007;
  (ans *= 1 + zero_cnt) %= 1000000007;
  cout << ans << endl;
  return 0;
}
