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;
char s[1000010];
const int mod = 1e9 + 7;
int dp[1000010], sum;
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> s;
  int n = strlen(s);
  int l = 0;
  while (s[l] == '0') l++;
  int ans = l + 1;
  int r = n - 1;
  while (r >= l && s[r] == '0') r--;
  ans = 1ll * ans * (n - r) % mod;
  if (l <= r) {
    sum = 1;
    for (int i = l + 1; i <= r;) {
      int j = i;
      while (j < r && s[j] == '0') j++;
      int x = j - i;
      int prv = sum;
      for (int k = 0; k <= x; k++) {
        sum = (sum - dp[k]) % mod;
        dp[k] = prv;
        sum = (sum + dp[k]) % mod;
      }
      if (sum < 0) sum += mod;
      i = j + 1;
    }
    ans = 1ll * ans * sum % mod;
  } else {
    ans--;
    if (ans < 0) ans += mod;
  }
  cout << ans;
}