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;
int mod = 1e9 + 7;
char c[1000005];
long long dp[1000005];
int main() {
  scanf("%s", c);
  vector<long long> v;
  int cnt = 0;
  for (int i = 0; c[i] != 0; i++) {
    if (c[i] == '0')
      cnt++;
    else {
      v.push_back(cnt + 1);
      cnt = 0;
    }
  }
  v.push_back(cnt + 1);
  if (v.size() == 1) {
    printf("%d\n", v[0] - 1);
  } else if (v.size() == 2) {
    printf("%lld\n", v[0] * v[1] % mod);
  } else {
    long long sum = v[1] + 1;
    fill(dp, dp + v[1] + 1, 1);
    for (int i = 2; i < v.size() - 1; i++) {
      long long sum2 = sum;
      for (int j = 1; j <= v[i]; j++) {
        sum2 += sum - dp[j];
        dp[j] += sum - dp[j];
        sum2 = (sum2 % mod + mod) % mod;
        dp[j] = (dp[j] % mod + mod) % mod;
      }
      sum = sum2;
    }
    printf("%lld\n", (sum)*v[0] % mod * v.back() % mod);
  }
}
