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 std::cerr;
const int N = 1000005, mu = 1e9 + 7;
char s[N];
int a[N], n, lst[N], pre[N], dp[N];
int main() {
  scanf("%s", s + 1);
  int m = strlen(s + 1), cnt = 0;
  for (int i = 1; i <= m; i++) {
    if (s[i] == '1')
      a[++n] = cnt, cnt = 0;
    else
      cnt++;
  }
  a[++n] = cnt;
  if (cnt == m) return printf("%d\n", m), 0;
  dp[1] = pre[1] = 1;
  for (int i = 0; i <= m; i++) lst[i] = 1;
  for (int i = 2; i < n; i++) {
    for (int j = 0; j <= a[i]; j++)
      dp[i] = ((long long)dp[i] + pre[i - 1] - pre[lst[j] - 1] + mu) % mu;
    pre[i] = (pre[i - 1] + dp[i]) % mu;
    for (int j = 0; j <= a[i]; j++) lst[j] = i;
  }
  int ans = pre[n - 1];
  ans = (long long)ans * (a[1] + 1) % mu * (a[n] + 1) % mu;
  printf("%d\n", ans);
}
