Problem: 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.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int mod = 1e9 + 7;
int main() {
  string s;
  cin >> s;
  int pref = 0, suf = 0;
  while (pref < s.size() and s[pref] == '0') pref++;
  while (suf < s.size() and s[s.size() - 1 - suf] == '0') suf++;
  if (pref == s.size())
    cout << s.size();
  else {
    s.erase(s.begin(), s.begin() + pref);
    s.erase(s.end() - suf, s.end());
    string s2 = s;
    int n = s.size(), pr = -1, ne[n];
    vector<pair<int, int> > st;
    while (s.size()) {
      char c = s.back();
      int q = 0;
      while (s.size() and s.back() == c) s.pop_back(), q++;
      if (c == '0') {
        for (int a = s.size(); a < s.size() + q - 1; a++) {
          if (s.size() + q < n)
            ne[a] = s.size() + q;
          else
            ne[a] = -1;
        }
        while (st.size() and st.back().first <= q) st.pop_back();
        if (st.size())
          ne[s.size() + q - 1] = st.back().second + q;
        else
          ne[s.size() + q - 1] = -1;
        st.push_back({q, s.size()});
      } else {
        for (int a = s.size(); a < s.size() + q - 1; a++) {
          if (s.size() + q < n)
            ne[a] = s.size() + q;
          else
            ne[a] = -1;
        }
        ne[s.size() + q - 1] = pr, pr = s.size();
      }
    }
    int dp[n], an = 0;
    for (int a = 0; a < n; a++) dp[a] = 0;
    dp[0] = 1;
    for (int a = 0; a + 1 < n; a++) {
      dp[a + 1] = (dp[a + 1] + dp[a]) % mod;
      if (ne[a] != -1) dp[ne[a]] = (dp[ne[a]] + dp[a]) % mod;
    }
    for (int a = 0; a < n; a++) {
      if (s2[a] == '1') an = (an + dp[a]) % mod;
    }
    an = 1ll * an * (pref + 1) % mod;
    an = 1ll * an * (suf + 1) % mod;
    cout << an;
  }
}