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;
const int INF = 0x7fffffff;
const int N = 1e6 + 5;
inline int read() {
  int x = 0, rev = 0, ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') rev = 1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = (x << 1) + (x << 3) + ch - '0';
    ch = getchar();
  }
  return rev ? -x : x;
}
char s[N];
int n, ans, flg, f[N], lst[N], p0[N], p1[N], pos[N];
int main() {
  cin >> s + 1;
  n = strlen(s + 1);
  for (int i = 1; i <= n && !flg; i++)
    if (s[i] == '1') {
      f[flg = i] = i;
    }
  if (!flg) return cout << n << endl, 0;
  for (int i = 1; i <= n; i++) {
    pos[i] = n + 1, lst[i] = (s[i] == '0') ? lst[i - 1] + 1 : 0;
  }
  for (int i = n; i; pos[lst[i]] = i, i--) {
    p0[i] = pos[lst[i] + 1], p1[i] = pos[0];
  }
  for (int i = flg; i < n; i++)
    (f[p0[i]] += f[i]) %= 1000000007, (f[p1[i]] += f[i]) %= 1000000007;
  for (; flg <= n; ++flg) (ans += f[flg] * (lst[flg] <= lst[n])) %= 1000000007;
  cout << ans << endl;
  return 0;
}
