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 = 1000000007;
int main() {
  string s;
  cin >> s;
  int n = s.length();
  vector<int> dist(n);
  for (int i = 0; i < n; i++)
    if (s[i] == '0') dist[i] = (i ? dist[i - 1] : 0) + 1;
  vector<int> dp(n + 2), nxt(n + 2, n);
  auto get = [&](int i) { return nxt[i] < n ? dp[nxt[i]] : 0; };
  for (int i = n - 1; i >= 0; i--) {
    dp[i] = ((dist[i] <= dist.back()) + get(0) + get(dist[i] + 1)) % mod;
    nxt[dist[i]] = i;
  }
  int ans = n;
  if (nxt[0] < n) ans = ((long long)get(0) * (nxt[0] + 1)) % mod;
  cout << ans << '\n';
  return 0;
}