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 long long MOD = 1000000007;
char st[1000002];
int L, N, a[1000001], q[1000001], D;
long long BO, f[1000001];
int main() {
  scanf("%s", st + 1);
  while (st[L + 1]) L++;
  for (int i = 1; i <= L; i++)
    if (st[i] == '1') a[++N] = i;
  if (N == 0) {
    printf("%d\n", L);
    return 0;
  }
  BO = (long long)a[1] * (L - a[N] + 1) % MOD;
  for (int i = 1; i < N; i++) a[i] = a[i + 1] - a[i];
  a[N--] = 0;
  f[0] = 1;
  long long S = 1;
  for (int i = 1; i <= N; i++) {
    while (D && a[i] > a[q[D]]) {
      S = (S + ((D > 1 ? f[q[D - 1] - 1] : 0) - f[q[D] - 1] + MOD) * a[q[D]]) %
          MOD;
      D--;
    }
    q[++D] = i;
    f[i] = S =
        (S + (f[i - 1] - (D > 1 ? f[q[D - 1] - 1] : 0) + MOD) * a[i]) % MOD;
  }
  printf("%lld\n", f[N] * BO % MOD);
  return 0;
}
