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;
mt19937 rnd(chrono::high_resolution_clock::now().time_since_epoch().count());
const int M = 1e9 + 7;
void add(int &a, int b) {
  a += b;
  if (a >= M) a -= M;
}
int mul(int a, int b) { return (a * (long long)b) % M; }
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  string s;
  cin >> s;
  int n = (int)s.size();
  map<char, int> last;
  last['0'] = n;
  last['1'] = n;
  vector<int> j0(n), j1(n);
  for (int i = n - 1; i >= 0; i--) {
    last[s[i]] = i;
    j0[i] = last['0'];
    j1[i] = last['1'];
  }
  int i = 0;
  vector<int> ok;
  vector<int> ch;
  while (i < n) {
    ch.push_back(s[i] - '0');
    int j = i;
    int len = 0;
    while (j < n && s[i] == s[j]) {
      j++;
      len++;
    }
    i = j;
    ok.push_back(len);
  }
  vector<int> one((int)ok.size() + 1), zero((int)ok.size() + 1);
  one[0] = 1, zero[0] = 1;
  int m = (int)ok.size();
  int sum = 0;
  vector<int> push((int)ok.size());
  vector<int> append((int)ok.size());
  vector<int> first_more(ok.size());
  vector<int> st;
  for (int i = (int)ok.size() - 1; i >= 0; i--) {
    if (ch[i] == 0) {
      while (!st.empty() && ok[st.back()] <= ok[i]) {
        st.pop_back();
      }
      if (st.empty())
        first_more[i] = m;
      else
        first_more[i] = st.back();
      st.push_back(i);
    }
  }
  if (ch[0] == 1) one[0] = 0;
  for (int i = 0; i < m; i++) {
    int mx = 0;
    if (ch.back() == 0 && (i == m - 1 || i != 0)) {
      add(zero[m], mul(one[i], ok.back()));
    }
    for (int j = i; j + 1 < m; j++) {
      if (ch[j] == 0) {
        if (i != 0) {
          add(append[j], mul(ok[j], one[i]));
          add(push[j], one[i]);
        } else {
          add(zero[j + 1], mul(ok[j], one[i]));
        }
        break;
      }
    }
    if (ch[i] == 0 && first_more[i] < m) {
      add(append[first_more[i]], mul(push[i], ok[first_more[i]] - ok[i]));
      add(push[first_more[i]], push[i]);
    }
    if (ch[i] == 0 && i + 1 < m) {
      add(zero[i + 1], append[i]);
    }
    add(sum, zero[i]);
    if (ch[i] == 1) {
      add(one[i + 1], mul(sum, ok[i]));
    }
  }
  int ans = zero[m];
  for (int i = 1; i <= m; i++) {
    add(ans, one[i]);
  }
  cout << ans << endl;
}