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;
string s;
long long ans, all1, all1full, bonus, All1[1000006];
vector<pair<long long, long long> > v;
stack<pair<int, long long> > zeroes;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cin >> s;
  long long sz = s.size(), l = 0;
  while (l < sz && s[l] == '0') l++;
  if (l == sz) {
    cout << sz << '\n';
    return 0;
  }
  while (s[sz - 1] == '0') sz--;
  bonus = (long long)(l + 1) * (long long)(s.size() + 1 - sz) % 1000000007;
  int X = sz - 1;
  while (X >= 0 && s[X] == '1') X--;
  ans = sz - X - 1;
  all1 = ans;
  all1full = 1;
  for (int i = X; i >= l;) {
    int j = i;
    while (s[j] == '0') j--;
    int jj = j;
    while (s[jj] == '1') jj--;
    v.push_back({j - jj, i - j});
    i = jj;
  }
  int n = v.size();
  for (int i = 0; i < n; i++) {
    All1[i] = all1;
    if (i > 0) All1[i] += All1[i - 1];
    All1[i] %= 1000000007;
    long long ALL1 = all1;
    all1 = all1 * v[i].second % 1000000007 * v[i].first + all1full * v[i].first;
    all1 %= 1000000007;
    ans += all1;
    ans %= 1000000007;
    all1full = ALL1 * v[i].second + all1full;
    all1full %= 1000000007;
    long long sum = 0;
    while (!zeroes.empty() && zeroes.top().second < v[i].second) {
      auto p = zeroes.top();
      zeroes.pop();
      long long curr;
      if (zeroes.empty())
        curr = All1[p.first];
      else
        curr = All1[p.first] - All1[zeroes.top().first];
      if (curr < 0) curr += 1000000007;
      curr *= v[i].second - p.second;
      sum = (sum + curr) % 1000000007;
    }
    if (!zeroes.empty() && zeroes.top().second == v[i].second) zeroes.pop();
    zeroes.push({i, v[i].second});
    all1full += sum;
    all1full %= 1000000007;
    all1 += sum * v[i].first;
    all1 %= 1000000007;
    ans += sum * v[i].first;
    ans %= 1000000007;
  }
  ans *= bonus;
  cout << ans % 1000000007 << '\n';
}