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;
template <typename G1, typename G2 = G1, typename G3 = G1>
struct triple {
  G1 first;
  G2 second;
  G3 T;
};
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  string s;
  cin >> s;
  int n = s.size();
  vector<int> p(n);
  for (int i = 0; i < n; i++)
    if (s[i] == '0') p[i] = (i == 0 ? 0 : p[i - 1]) + 1;
  const int mod = 1e9 + 7;
  vector<int> l(n + 2, n), dp(n);
  for (int i = n - 1; i >= 0; i--) {
    dp[i] = (p[i] <= p[n - 1]);
    dp[i] += (l[0] == n ? 0 : dp[l[0]]);
    dp[i] += (l[p[i] + 1] == n ? 0 : dp[l[p[i] + 1]]);
    dp[i] %= mod;
    l[p[i]] = i;
  }
  int k = 0;
  for (; k < n; k++)
    if (s[k] == '1') break;
  cout << (k == n ? n : (long long)dp[k] * (k + 1) % mod) << '\n';
  return 0;
}