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;
int n, l, r, a[1000005], nxt[1000005], cnt;
long long ans, mul, dp[1000005], mod = 1e9 + 7;
string s;
int main() {
  cin >> s;
  n = s.length();
  cnt = 1;
  while (s.size() && s.back() == '0') {
    cnt++;
    s.pop_back();
  }
  mul = cnt;
  int ptr = 0;
  if (!s.size()) {
    cout << mul - 1 << "\n";
    return 0;
  }
  while (ptr < s.length() && s[ptr] == '0') {
    ptr++;
  }
  mul *= (ptr + 1);
  mul %= mod;
  s = s.substr(ptr);
  for (int i = 1; i < s.length(); i++) {
    if (s[i] == '0')
      a[i] = a[i - 1] + 1;
    else
      a[i] = 0;
  }
  dp[s.length() - 1] = 1;
  nxt[0] = s.length() - 1;
  for (int i = (int)s.length() - 2; i >= 0; i--) {
    dp[i] = dp[nxt[a[i] + 1]] + dp[nxt[0]] + (s[i] == '1');
    dp[i] %= mod;
    nxt[a[i]] = i;
  }
  ans = dp[0] * mul % mod;
  cout << ans << "\n";
}
