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;
char s[1000010];
int len;
int last;
const int mod = 1e9 + 7;
int f[1000010];
int Op[1000010];
int ne[1000010];
int st[1000010], num;
int Hou[1000010], Qian[1000010];
int main() {
  scanf("%s", &s[1]);
  len = strlen(&s[1]);
  for (int i = len; i >= 1; i--) {
    Op[i] = last;
    if (s[i] == '1') last = i;
  }
  if (!last)
    cout << len;
  else {
    int start;
    for (int i = 1; i <= len; i++)
      if (s[i] == '1') {
        f[i] = i;
        start = i;
        break;
      }
    for (int i = 1; i <= len; i++)
      if (s[i] == '0') Qian[i] = Qian[i - 1] + 1;
    for (int i = len; i >= 1; i--)
      if (s[i] == '0') Hou[i] = 1 + Hou[i + 1];
    if (s[len] == '0') st[num = 1] = len - Qian[len] + 1;
    for (int i = len; i >= 1; i--)
      if (s[i] == '0' && s[i + 1] == '1') {
        while (num && Hou[st[num]] <= Qian[i]) num--;
        if (!num)
          ne[i] = -1;
        else
          ne[i] = st[num] + Qian[i];
        st[++num] = i - Qian[i] + 1;
      }
    for (int i = len; i >= 1; i--)
      if (s[i] == '1') {
        if (s[i + 1] == '0')
          ne[i] = i + 1;
        else
          ne[i] = ne[i + 1];
      }
    for (int i = start; i <= len; i++) {
      f[Op[i]] = (f[Op[i]] + f[i]) % mod;
      if (s[i + 1] == '0')
        f[i + 1] = (f[i + 1] + f[i]) % mod;
      else if (ne[i] != -1)
        f[ne[i]] = (f[ne[i]] + f[i]) % mod;
    }
    int ans = 0;
    int Fd = Qian[len];
    for (int i = 1; i <= len; i++)
      if (s[i] == '1')
        ans = (ans + f[i]) % mod;
      else if (Qian[i] <= Fd)
        ans = (ans + f[i]) % mod;
    cout << ans << '\n';
  }
  return 0;
}
