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 mod = 1e9 + 7;
char s[1000005];
int dp[1000005], sum[1000005];
int main() {
  scanf("%s", s + 1);
  int n = strlen(s + 1);
  vector<int> v;
  for (int i = 1; i <= n; i++) {
    if (s[i] == '1') v.emplace_back(i);
  }
  if (v.size() == 0) {
    printf("%d\n", n);
    return 0;
  }
  dp[0] = 1;
  sum[0] = 1;
  int tmp = 1, sum2 = 0;
  ;
  for (int i = 1; i < v.size(); i++) {
    int l = v[i] - v[i - 1] - 1;
    for (int j = 0; j <= l; j++) {
      tmp = (tmp - sum[j] + mod) % mod;
      sum[j] = dp[i - 1];
      tmp = (tmp + sum[j]) % mod;
    }
    dp[i] = (tmp + 1) % mod;
  }
  printf("%d\n",
         dp[v.size() - 1] % mod * 1ll * v[0] % mod * (n - v.back() + 1) % mod);
}
