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 uint64_t mod = 1000000007;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  string s;
  cin >> s;
  uint64_t start = 0, end = s.size();
  while (start < end && s[start] == '0') {
    start++;
  }
  if (start == end) {
    cout << s.size() << endl;
    return 0;
  }
  while (s[end - 1] == '0') {
    end--;
  }
  vector<uint64_t> a;
  uint64_t az, al;
  az = start;
  al = s.size() - end;
  for (int prev = start, i = start + 1; i < end; ++i) {
    if (s[i] == '1') {
      a.push_back(i - prev - 1);
      prev = i;
    }
  }
  int n = a.size();
  uint64_t prev_dps = 1;
  vector<uint64_t> prev_dpa(s.size() + 1);
  for (int i = 0; i < n; ++i) {
    vector<uint64_t> dp(a[i] + 1);
    uint64_t new_dps = prev_dps;
    for (uint64_t x = 0; x <= a[i]; ++x) {
      dp[x] = (prev_dps - prev_dpa[x] + mod) % mod;
      new_dps = (new_dps + dp[x]) % mod;
      prev_dpa[x] = (prev_dpa[x] + dp[x]) % mod;
    }
    prev_dps = new_dps;
  }
  cout << (prev_dps * (((az + 1) * (al + 1)) % mod)) % mod << endl;
}
