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 int nmax = 1e6 + 42, mod = 1e9 + 7;
string inp;
int n;
long long dp[nmax];
int solve(int l, int r) {
  vector<int> SZ = {};
  for (int i = l; i < r; i++) {
    int j = i + 1;
    while (j < r && inp[j] == '0') j++;
    SZ.push_back(j - i - 1);
    i = j - 1;
  }
  long long sum = 1;
  for (auto k : SZ) {
    long long change = 0;
    for (int j = 0; j <= k; j++) {
      change = (change + sum - dp[j] + mod) % mod;
      dp[j] = sum;
    }
    sum = (sum + change) % mod;
  }
  return sum;
}
int main() {
  cin >> inp;
  n = inp.size();
  int lhs = 0, rhs = n - 1;
  while (lhs < n && inp[lhs] == '0') lhs++;
  while (rhs >= 0 && inp[rhs] == '0') rhs--;
  if (lhs > rhs) {
    cout << n << endl;
    return 0;
  }
  long long outp = (lhs + 1);
  outp = outp * (n - rhs) % mod;
  outp = outp * solve(lhs, rhs) % mod;
  cout << outp << endl;
  return 0;
}
