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 mod = 1e9 + 7;
void add(int& a, const int& b) {
  a += b;
  if (a >= mod) a -= mod;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  string s;
  cin >> s;
  int suf0 = 0;
  while (!s.empty() && s.back() == '0') {
    s.pop_back();
    suf0++;
  }
  if (s.empty()) {
    cout << suf0 << "\n";
    return 0;
  }
  int n = s.size();
  vector<int> z(n, 0);
  for (int i = 0; i < n; i++) {
    if (s[i] == '0') {
      if (!i)
        z[i] = 1;
      else
        z[i] = z[i - 1] + 1;
    }
  }
  vector<int> dp(n + 1, 0), nxt(n + 1, n);
  for (int i = n - 1; i >= 0; i--) {
    if (s[i] == '1') dp[i] = 1;
    add(dp[i], dp[nxt[0]]);
    add(dp[i], dp[nxt[z[i] + 1]]);
    nxt[z[i]] = i;
  }
  for (int i = 0; i < n; i++) {
    if (s[i] == '1') {
      cout << (((dp[i] * 1ll * (suf0 + 1)) % mod) * 1ll * (i + 1)) % mod
           << "\n";
      return 0;
    }
  }
  return 0;
}