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 int N_MAX = 1e6, MOD = 1e9 + 7;
int n, dp[N_MAX], sum;
string s;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cin >> s;
  n = s.length();
  int i = 0;
  while (i < n && s[i] == '0') i++;
  if (i == n) {
    cout << n << "\n";
    return 0;
  }
  int j = n - 1;
  while (s[j] == '0') j--;
  long long mult = i + 1;
  mult *= n - j;
  mult %= MOD;
  s = s.substr(i, j - i + 1);
  n = s.length();
  sum = 1;
  i = 0;
  while (i < n - 1) {
    int j = 0;
    while (i + j + 1 < n && s[i + j + 1] == '0') j++;
    int diff = 0;
    for (int k = 0; k <= j; k++) {
      diff += sum;
      if (diff >= MOD) diff -= MOD;
      diff += MOD - dp[k];
      if (diff >= MOD) diff -= MOD;
      dp[k] = sum;
    }
    sum += diff;
    if (sum >= MOD) sum -= MOD;
    i += j + 1;
  }
  cout << mult * sum % MOD << "\n";
  return 0;
}
