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;
using ll = long long;
const int P = int(1e9) + 7;
signed main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  string str;
  cin >> str;
  int n = int((str).size());
  if (str == string(n, '0')) {
    cout << n << '\n';
    return 0;
  }
  int s = 0;
  while (str[s] == '0') s++;
  int f = n - 1;
  while (str[f] == '0') f--;
  vector<int> zeroes;
  for (int i = int(s); i < int(f); i++) {
    assert(str[i] == '1');
    int j = i + 1;
    while (str[j] == '0') j++;
    j--;
    zeroes.push_back(j - i);
    i = j;
  }
  int zero = int((zeroes).size()) + 1;
  vector<int> dp(zero);
  dp[0] = 1;
  vector<int> pref{0, 1};
  for (int i = int(1); i < int(zero); i++) {
    dp[i] = 1ll * dp[i - 1] * (zeroes[i - 1] + 1) % P;
    int sufMax = INT_MIN;
    for (int j = i - 2; j >= 0; j--) {
      sufMax = max(sufMax, zeroes[j]);
      if (sufMax >= zeroes[i - 1]) break;
      dp[i] += 1ll * (zeroes[i - 1] - sufMax) * dp[j] % P;
      if (dp[i] < 0) dp[i] += P;
      if (dp[i] >= P) dp[i] -= P;
    }
    pref.push_back((pref.back() + dp[i]) % P);
  }
  cout << 1ll * pref.back() * (s + 1) % P * (n - f) % P << '\n';
}
