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 = int(1e6) + 5;
const int mo = int(1e9) + 7;
int n;
char s[N];
int cnt0[N], nexcnt0[N], f[N];
int main() {
  scanf("%s", s + 1);
  n = strlen(s + 1);
  for (int i = 1; i <= n; i++) s[i] -= '0';
  for (int i = 1; i <= n; i++)
    if (s[i] == 0) cnt0[i] = cnt0[i - 1] + 1;
  for (int i = 0; i <= n + 1; i++) nexcnt0[i] = n + 1;
  for (int i = n; i >= 1; i--) {
    f[i] =
        ((cnt0[i] <= cnt0[n]) + f[nexcnt0[cnt0[i] + 1]] + f[nexcnt0[0]]) % mo;
    nexcnt0[cnt0[i]] = i;
  }
  printf("%d\n",
         nexcnt0[0] <= n ? int((long long)f[nexcnt0[0]] * nexcnt0[0] % mo) : n);
}