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;
int nxt[1000100][2];
long long way[1000100];
const int mod = 1000000007;
signed main() {
  std::ios::sync_with_stdio(false);
  string s;
  cin >> s;
  int n = s.size();
  int a1 = 0, a2 = 0;
  for (int i = 0; i < n; ++i) {
    if (s[i] == '0')
      ++a1;
    else {
      break;
    }
  }
  for (int i = n - 1; i >= 0; --i) {
    if (s[i] == '0')
      ++a2;
    else {
      break;
    }
  }
  if (a1 == s.size()) {
    cout << n << '\n';
    return 0;
  }
  s = s.substr(a1, n - a1 - a2);
  n = s.size();
  for (int i = n - 1; i >= 0; --i) {
    for (char c = 0; c <= 1; ++c)
      if (s[i] == c + '0')
        nxt[i][c] = i + 1;
      else
        nxt[i][c] = nxt[i + 1][c];
  }
  vector<pair<int, int>> v;
  int t = 0, q = s.size() - 1;
  for (int i = s.size() - 1; i >= 0; --i) {
    if (s[i] == '0') {
      q = i;
      ++t;
    } else if (t) {
      while (v.size() && v.back().second <= t) v.pop_back();
      if (v.size())
        nxt[q + t][0] = v.back().first + t + 1;
      else
        nxt[q + t][0] = 0;
      v.push_back({q, t});
      t = 0;
    }
  }
  long long ans = 0;
  way[1] = 1;
  for (int i = 0; i <= n; ++i) {
    if (nxt[i][0]) (way[nxt[i][0]] += way[i]) %= mod;
    if (nxt[i][1]) (way[nxt[i][1]] += way[i]) %= mod;
    if (s[i - 1] == '1') (ans += way[i]) %= mod;
  }
  cout << ans * (a1 + 1) % mod * (a2 + 1) % mod << '\n';
  return 0;
}
