Problem: 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.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const long long int N = 1e6 + 5, inf = 1e18, mod = 1e9 + 7;
long long int freq[N];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  string s;
  cin >> s;
  long long int n = ((long long int)((s).size()));
  long long int c = 0;
  long long int ans = 1;
  long long int flag = 0;
  for (long long int i = n - 1; i >= 0; i--) {
    if (s[i] == '0') {
      c++;
    } else {
      if (flag == 0) {
        ans = c + 1;
        c = 0;
        flag = 1;
      } else {
        long long int cur = (ans * (c + 1)) % mod;
        long long int pre = ans;
        for (long long int i = 0; i < c + 1; ++i) {
          pre -= freq[i];
          pre += mod;
          pre %= mod;
          freq[i] = ans;
        }
        c = 0;
        ans = cur + pre;
        ans %= mod;
      }
    }
  }
  ans = (ans * min(n, (c + 1))) % mod;
  cout << ans << "\n";
  return 0;
}