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 gcd(long long a, long long b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
long long dp1[1000010], dp2[1000010];
long long add(long long a, long long b) {
  return (a + b >= ((long long)1e9 + 7) ? a + b - ((long long)1e9 + 7) : a + b);
}
long long sub(long long a, long long b) {
  return (a < b ? a + ((long long)1e9 + 7) - b : a - b);
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  string s;
  cin >> s;
  long long l = 0, r = s.length() - 1;
  bool tag = false;
  for (long long i = 0; i < s.length(); i++)
    if (s[i] == '1') {
      tag = true;
      break;
    }
  if (!tag) {
    cout << s.length() << '\n';
    return 0;
  }
  while (s[l] == '0') l++;
  while (s[r] == '0') r--;
  long long cnt = 0, ss = 0, rem = 0, ct = 0, sum = 0;
  memset(dp1, 0, sizeof(dp1));
  for (long long i = l; i <= r; i++) {
    if (s[i] == '0') {
      cnt++;
      ss = sub(ss, dp1[cnt]);
      ss = add(ss, rem);
      dp1[cnt] = rem;
    } else {
      cnt = 0;
      ct++;
      sum = add(sum, ss);
      rem = add(sum, ct);
      dp2[i] = rem;
    }
  }
  cout << (((rem * (l + 1)) % ((long long)1e9 + 7)) * (s.length() - r)) %
              ((long long)1e9 + 7)
       << '\n';
  return 0;
}
