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;
#pragma GCC optimize("O2,no-stack-protector,unroll-loops,fast-math")
const long long maxn = 1e6 + 10, maxm = 5e4 + 10, lg = 21, mod = 1e9 + 7,
                inf = 1e18;
string s;
long long cc[maxn], t0, las, sig, ff = -1;
int main() {
  ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  cin >> s;
  for (int i = (long long)(s.size()) - 1; i >= 0; i--) {
    if (s[i] == '1') {
      if (ff == -1) ff = t0;
      for (int j = 0; j <= t0; j++)
        sig -= cc[j], sig += mod, sig += cc[j] = (las + (j <= ff)) % mod,
                                         sig %= mod;
      t0 = 0, las = sig;
    } else
      t0++;
  }
  cout << (t0 == (long long)(s.size()) ? t0 : (t0 + 1) * sig % mod);
  return 0;
}
