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;
template <class T>
bool ckmin(T& a, const T& b) {
  return b < a ? a = b, 1 : 0;
}
template <class T>
bool ckmax(T& a, const T& b) {
  return a < b ? a = b, 1 : 0;
}
const long long MOD = (long long)(1e9) + 7;
long long getVal(pair<pair<int, int>, long long> p) {
  return (p.second * (p.first.second - p.first.first + 1)) % MOD;
}
void pr(stack<pair<pair<int, int>, long long>> st) {
  cout << st.top().first.first << " " << st.top().first.second << " "
       << st.top().second << "\n";
  st.pop();
  if (!st.empty()) pr(st);
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  string second;
  cin >> second;
  vector<int> v;
  int cur = 0;
  for (int i = 0; i < (int)second.size(); i++) {
    if (second[i] == '1') {
      v.push_back(cur);
      cur = 0;
    } else {
      cur++;
    }
  }
  if ((int)v.size() == 0) {
    return cout << (int)second.size() << '\n', 0;
    ;
  }
  long long factor = ((long long)(v[0] + 1) * (cur + 1)) % MOD;
  v.erase(v.begin());
  stack<pair<pair<int, int>, long long>> st;
  long long cursum = 0;
  for (int i = 0; i < (int)v.size(); i++) {
    long long ori = cursum;
    while (!st.empty()) {
      if (v[i] >= st.top().first.first) {
        cursum -= getVal(st.top());
        cursum %= MOD;
        cursum += MOD;
        cursum %= MOD;
        if (v[i] < st.top().first.second) {
          st.top().first.first = v[i] + 1;
          cursum += getVal(st.top());
          cursum %= MOD;
          break;
        } else {
          st.pop();
        }
      } else {
        break;
      }
    }
    st.push(make_pair(make_pair(0, v[i]), (ori + 1) % MOD));
    cursum += getVal(st.top());
    cursum %= MOD;
  }
  cursum++;
  cursum %= MOD;
  long long res = (factor * cursum) % MOD;
  return cout << res << '\n', 0;
  ;
}
