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>
const int MOD = 1e9 + 7;
char buffer[1000006];
char* str = buffer;
int N;
int lead[1000006];
int dp[1000006];
int dp1sum[1000006];
int recent[1000006];
int solve() {
  assert(str[0] == '1');
  dp[0] = 1;
  dp1sum[0] = 1;
  int zero_run = 0;
  for (int i = 1; i < N; i++) {
    if (str[i] == '1') {
      lead[i] = zero_run;
      for (int k = 1; k <= zero_run; k++) {
        recent[k] = i;
      }
      zero_run = 0;
      int j = i;
      do {
        j--;
        dp[i] = (dp[i] + dp[j]) % MOD;
      } while (str[j] != '1');
      dp1sum[i] = (dp1sum[i - 1] + dp[i]) % MOD;
    } else {
      zero_run++;
      dp[i] = ((dp1sum[i - 1] - dp1sum[recent[zero_run]] + MOD) % MOD +
               dp[recent[zero_run]]) %
              MOD;
      dp1sum[i] = dp1sum[i - 1];
    }
  }
  return dp1sum[N - 1];
}
int main() {
  scanf("%s", buffer);
  int head = 0, tail = 0;
  while (*str == '0') {
    head++;
    str++;
  }
  if (*str == 0) {
    printf("%d\n", head);
    return 0;
  }
  N = strlen(str);
  while (str[N - 1] == '0') {
    tail++;
    N--;
  }
  assert(str[N - 1] == '1');
  str[N] = '\0';
  printf("%d\n", int(solve() * (head + 1LL) % MOD * (tail + 1LL) % MOD));
}
