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;
int n, m, pre[1000005];
int w[1000005], dp[1000005], sum[1000005], A[1000005];
char S[1000005];
signed main() {
  scanf("%s", S + 1);
  n = strlen(S + 1);
  for (int i = 1; i <= n; i++) A[i] = S[i] - '0';
  int num = 0;
  for (int i = 1; i <= n; i++) {
    if (A[i] == 1)
      pre[++m] = num, num = 0;
    else
      num++;
  }
  pre[++m] = num;
  if (m == 1) {
    printf("%d\n", n);
    return 0;
  }
  int ans = (long long)(pre[1] + 1) * (pre[m] + 1) % 1000000007;
  for (int i = 0; i <= n; i++) w[i] = 1;
  sum[1] = dp[1] = 1;
  for (int i = 2; i < m; i++) {
    int p = w[0];
    for (int j = 0; j <= pre[i]; j++) {
      p = w[j];
      dp[i] = ((dp[i] + sum[i - 1]) % 1000000007 - sum[p - 1] + 1000000007) %
              1000000007;
    }
    for (int j = 0; j <= pre[i]; j++) w[j] = i;
    sum[i] = (sum[i - 1] + dp[i]) % 1000000007;
  }
  ans = (long long)ans * sum[m - 1] % 1000000007;
  printf("%d\n", ans);
}