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 N = 1E6 + 5, MOD = 1E9 + 7;
int cnt = 1;
long long ans = 1, cur = 1;
bool st = true;
vector<pair<int, long long>> ve = {{N, 0}};
string s;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cin >> s;
  for (int i = 0;; i++) {
    if (i == s.size()) {
      ans = (ans * cnt % MOD * cur - st) % MOD;
      return cout << (ans + MOD) % MOD, 0;
    }
    if (s[i] == '0') {
      cnt++;
    } else {
      if (st) {
        ans = cnt;
        st = false;
      } else {
        long long lst = cur;
        int tmp = cnt;
        while (ve.back().first <= tmp) {
          (cur -= ve.back().first * ve.back().second) %= MOD;
          tmp -= ve.back().first;
          ve.pop_back();
        }
        (cur -= tmp * ve.back().second) %= MOD;
        ve.back().first -= tmp;
        (cur += cnt * lst) %= MOD;
        ve.push_back({cnt, lst});
      }
      cnt = 1;
    }
  }
}