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;
const int N = 1e6 + 10, skc = 1e9 + 7;
string s;
int a[N], b[N], n = 1, sum[N], f[N], nw[N];
void upd(int &x, int y) { x = (x + y) % skc; }
signed main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cin >> s;
  int l = s.length();
  s = ' ' + s;
  bool flag = 0;
  for (register int i = (1); i <= (l); ++i) flag |= (s[i] == '1');
  if (flag == 0) {
    cout << l << '\n';
    return 0;
  }
  for (register int i = (1); i <= (l); ++i)
    if (s[i] == '0')
      a[n]++;
    else
      n++;
  int ans = 1ll * (a[1] + 1) * (a[n] + 1) % skc;
  n--;
  sum[1] = f[1] = 1;
  for (register int i = (2); i <= (n); ++i) {
    for (register int j = (0); j <= (a[i]); ++j)
      upd(f[i], (sum[i - 1] - sum[nw[j] - 1] + skc) % skc);
    for (register int j = (0); j <= (a[i]); ++j) nw[j] = i;
    sum[i] = (sum[i - 1] + f[i]) % skc;
  }
  cout << 1ll * ans * sum[n] % skc << '\n';
  return 0;
}
