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;
int n, i, nr, sum, sol;
char s[1000005];
int nxt[1000005], d[1000005], r[1000005];
int modulo(int x) {
  if (x < 0) {
    return x + 1000000007;
  }
  if (x >= 1000000007) {
    return x - 1000000007;
  }
  return x;
}
int main() {
  cin >> s + 1;
  n = strlen(s + 1);
  nxt[n] = n + 1;
  for (i = n - 1; i >= 0; i--) {
    if (s[i + 1] == '1') {
      nxt[i] = i + 1;
    } else {
      nxt[i] = nxt[i + 1];
    }
  }
  if (nxt[0] == n + 1) {
    cout << n;
    return 0;
  }
  d[nxt[0]] = 1;
  for (i = nxt[0]; i <= n; i++) {
    if (s[i] == '1') {
      nr = 0;
      sum = modulo(sum + d[i]);
    } else {
      nr++;
      d[i] = modulo(sum - r[nr]);
      r[nr] = sum;
    }
    d[nxt[i]] = modulo(d[nxt[i]] + d[i]);
    if (s[i] == '1') {
      sol = modulo(sol + d[i]);
    }
  }
  sol = sol * 1LL * (nxt[0]) % 1000000007 * (nr + 1) % 1000000007;
  cout << sol;
}
