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 int MOD = 1e9 + 7;
void madd(int &x, int y) {
  if ((x += y) >= MOD) x -= MOD;
  if (x < 0) x += MOD;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(nullptr);
  string s;
  cin >> s;
  int ans = 1, n = s.size();
  vector<int> sum(n);
  if (count(s.begin(), s.end(), '0') == n) {
    cout << n << '\n';
    return 0;
  }
  int st = 0, en = n - 1;
  while (s[st] == '0') st++;
  while (s[en] == '0') en--;
  for (int i = st; i < en;) {
    int j = i + 1;
    while (s[j] == '0') j++;
    int cur = 0;
    for (int k = 0; k < j - i; k++) {
      madd(cur, ans - sum[k]);
      sum[k] = ans;
    }
    madd(ans, cur);
    i = j;
  }
  ans = 1ll * ans * (st + 1) % MOD;
  ans = 1ll * ans * (n - en) % MOD;
  cout << ans << '\n';
}