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 md = 1e9 + 7;
void add(int &a, int b) {
  a += b;
  if (a >= md) a -= md;
}
int mul(int a, int b) { return ((long long)a * b) % md; }
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  string s;
  cin >> s;
  int n = s.size();
  int cnt_l = 0;
  while (cnt_l < n && s[cnt_l] == '0') ++cnt_l;
  s = s.substr(cnt_l, n - cnt_l);
  n = s.size();
  if (n == 0) {
    cout << cnt_l << '\n';
    return 0;
  }
  int cnt_r = 0;
  while (cnt_r < n && s[n - cnt_r - 1] == '0') ++cnt_r;
  s = s.substr(0, n - cnt_r);
  n = s.size();
  vector<int> dp(n);
  vector<pair<int, int>> st;
  auto go_merge = [&]() {
    while (st.size() >= 2 && st.rbegin()[0].second == st.rbegin()[1].second) {
      add(st.rbegin()[1].first, st.rbegin()[0].first);
      st.pop_back();
    }
  };
  int sum = 0;
  int len = 0;
  for (int i = 0; i < n; ++i) {
    if (s[i] == '1') {
      add(dp[i], 1);
      add(dp[i], sum);
      len = 0;
      st.push_back({dp[i], 0});
    } else {
      ++len;
      if (st.back().second < len) {
        add(dp[i], st.back().first);
        st.back().second = len;
      }
      add(sum, dp[i]);
    }
    go_merge();
  }
  int ans = 0;
  for (int i = 0; i < n; ++i)
    if (s[i] == '1') add(ans, dp[i]);
  ans = mul(ans, mul(cnt_l + 1, cnt_r + 1));
  cout << ans << '\n';
}