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 long long mod = 1e9 + 7;
signed main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  string s;
  cin >> s;
  long long suff = 0;
  while (s.size() && s.back() == '0') suff++, s.pop_back();
  if (s.empty()) return cout << suff, 0;
  long long n = s.size();
  vector<long long> dis(n);
  for (long long i = 0; i < n; ++i) {
    if (s[i] == '0') dis[i] = (i ? dis[i - 1] : 0) + 1;
  }
  vector<long long> dp(n + 2), nxt(n + 2, n);
  auto get = [&](long long i) -> long long {
    return (nxt[i] < n ? dp[nxt[i]] : 0);
  };
  for (long long i = n - 1; ~i; --i) {
    dp[i] = (s[i] == '1') + get(0) + get(dis[i] + 1);
    dp[i] %= mod;
    nxt[dis[i]] = i;
  }
  long long ans = get(0) * (nxt[0] + 1) % mod;
  ans = ans * (suff + 1) % mod;
  cout << ans;
  return 0;
}
