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 N = 1e6 + 5, mo = 1e9 + 7;
int n, zero, ans;
int pre[N], f[N], pos[N], nx0[N], nx1[N];
char s[N];
int main() {
  scanf("%s", s + 1);
  n = strlen(s + 1);
  for (int i = 1, _e = n; i <= _e; ++i)
    if (s[i] == '1')
      break;
    else
      ++zero;
  if (zero == n) {
    printf("%d\n", n);
    return 0;
  }
  ++zero;
  for (int i = 1, _e = n; i <= _e; ++i)
    pre[i] = s[i] == '1' ? 0 : pre[i - 1] + 1;
  for (int i = n, _e = 1; i >= _e; --i) {
    pos[pre[i]] = i;
    nx0[i] = pos[pre[i] + 1];
    nx1[i] = s[i + 1] == '1' ? i + 1 : nx1[i + 1];
  }
  f[zero] = 1;
  for (int i = zero, _e = n; i <= _e; ++i) {
    f[nx0[i]] = (f[nx0[i]] + f[i]) % mo;
    f[nx1[i]] = (f[nx1[i]] + f[i]) % mo;
    if (pre[i] <= pre[n]) ans = (ans + f[i]) % mo;
  }
  printf("%d\n", (long long)ans * zero % mo);
}