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;
vector<vector<long long> > adj;
long long solve(string s) {
  long long son1 = -1;
  for (long long i = s.size() - 1; i >= 0; i--) {
    if (s[i] == '1') {
      if (son1 != -1) {
        adj[i].push_back(son1);
      }
      son1 = i;
    }
  }
  long long son0 = -1;
  for (long long i = s.size() - 1; i >= 0; i--) {
    if (s[i] == '0') {
      son0 = i;
    } else {
      if (son0 != -1) {
        adj[i].push_back(son0);
      }
    }
  }
  son1 = -1;
  for (long long i = s.size() - 1; i >= 0; i--) {
    if (s[i] == '1')
      son1 = i;
    else {
      if (son1 != -1) adj[i].push_back(son1);
    }
  }
  vector<long long> son0lar(s.size() + 1, -1);
  vector<long long> sayilar(s.size(), 0);
  long long sayac = 0;
  for (long long i = 0; i < s.size(); i++) {
    if (s[i] == '0') {
      sayac++;
      sayilar[i] = sayac;
    } else
      sayac = 0;
  }
  for (long long i = s.size() - 1; i >= 0; i--) {
    if (s[i] == '0') {
      son0lar[sayilar[i]] = i;
      if (son0lar[sayilar[i] + 1] != -1)
        adj[i].push_back(son0lar[sayilar[i] + 1]);
    }
  }
  vector<long long> degerler(s.size(), 0);
  degerler[0] = 1;
  long long ans = 0;
  for (long long i = 0; i < s.size(); i++) {
    for (long long j : adj[i]) {
      degerler[j] += degerler[i];
      degerler[j] %= 1000000007;
    }
    if (s[i] == '1') ans += degerler[i];
    ans %= 1000000007;
  }
  return ans;
}
int main(int argc, char* argv[]) {
  cin.tie(nullptr);
  cout.tie(nullptr);
  ios_base::sync_with_stdio(false);
  string s;
  cin >> s;
  adj.resize(s.size());
  long long ilk = 0;
  for (ilk = 0; ilk < s.size(); ilk++) {
    if (s[ilk] == '1') break;
  }
  long long son = s.size() - 1;
  for (; son >= 0; son--) {
    if (s[son] == '1') break;
  }
  if (ilk == s.size()) {
    cout << s.size() << endl;
  } else {
    long long ans = solve(s.substr(ilk, son + 1));
    ans *= ilk + 1;
    ans %= 1000000007;
    ans *= s.size() - son;
    ans %= 1000000007;
    cout << ans << endl;
  }
}