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 P = 1000000007;
const int maxn = 1000009;
int n;
char s[maxn];
int nex[maxn];
int le[maxn];
int f[maxn];
int main() {
  scanf("%s", s + 1);
  n = strlen(s + 1);
  int p = 0;
  for (int i = n; i >= 1; --i)
    if (s[i] == '1') p = i;
  if (!p) {
    printf("%d\n", n);
    return 0;
  }
  for (int i = 1; i <= n; ++i) {
    if (s[i] == '0')
      le[i] = le[i - 1] + 1;
    else
      le[i] = 0;
  }
  for (int i = n; i >= 1; --i) {
    if (le[i] <= le[n]) f[i] = 1;
    f[i] = (f[i] + f[nex[0]]) % P;
    f[i] = (f[i] + f[nex[le[i] + 1]]) % P;
    nex[le[i]] = i;
  }
  printf("%lld\n", 1LL * p * f[p] % P);
  return 0;
}
