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 c = 1000002;
string s;
long long n, dp[c], kom[c], lp[c], pos, sum = 1, e, db = 1, cnt,
                                        mod = 1000000007;
int main() {
  cin >> s, n = s.size();
  while (s[pos] == '0') pos++, dp[pos] = 1, sum++;
  if (pos == n) {
    cout << n;
    return 0;
  }
  while (pos < n) {
    pos++, dp[pos] = sum, kom[pos] = dp[pos] + kom[pos - 1], cnt = 0;
    while (pos < n && s[pos] == '0')
      cnt++, pos++, kom[pos] = kom[pos - 1],
                    dp[pos] = (kom[pos] - kom[lp[cnt]] + mod) % mod,
                    sum += dp[pos], sum %= mod, lp[cnt] = pos;
  }
  sum = 0;
  while (pos > 0) {
    if (s[pos - 1] == '1') e++, sum += dp[pos];
    if (!e) db++;
    pos--;
  }
  cout << sum * db % mod << "\n";
  return 0;
}