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 T, n;
char s[1000009];
int nu[1000009] = {0};
int dp[1000009] = {0};
int qz[1000009] = {0};
int l[1000009] = {0};
int main() {
  scanf("%s", s);
  int a, b, c, i, j, pr = 0, ans = 0, sum = 0;
  n = strlen(s);
  dp[0] = 1;
  qz[0] = 0;
  for (a = 0; a < n; a++) nu[a + 1] = s[a] - '0';
  for (a = 1; a <= n; a++) {
    if (nu[a] == 0) {
      for (b = a; nu[b] == 0 && b <= n; b++) {
        dp[b] = a == 1
                    ? 1
                    : (qz[a - 1] - qz[l[b - a + 1]] + 1000000007) % 1000000007;
        qz[b] = qz[b - 1];
      }
      for (c = 0; c <= b - a; c++) l[c] = b - 1;
      a = b - 1;
    } else {
      for (b = pr; b < a; b++) dp[a] = (dp[a] + dp[b]) % 1000000007;
      pr = a;
      qz[a] = (dp[a] + qz[a - 1]) % 1000000007;
    }
  }
  for (a = 1; a <= n; a++)
    if (nu[a] == 1) ans = (ans + dp[a]) % 1000000007;
  if (pr) {
    c = ans;
    for (a = pr + 1; a <= n; a++) ans = (c + ans) % 1000000007;
  } else
    ans = n;
  printf("%d\n", ans);
  return 0;
}