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 N = 1e6 + 5;
const int mod = 1e9 + 7;
int f[N], dst[N], pos[N], nxt1[N], nxtd[N];
char str[N];
int main() {
  scanf("%s", str + 1);
  int n = strlen(str + 1);
  int st = 0;
  for (int i = 1; i <= n; i++)
    if (str[i] == '1') {
      st = i;
      break;
    }
  if (!st)
    printf("%d\n", n);
  else {
    for (int i = 1; i <= n; i++)
      if (str[i] == '0') dst[i] = dst[i - 1] + 1;
    for (int i = 0; i <= n + 1; i++) pos[i] = n + 1;
    for (int i = n; i >= 1; i--) {
      nxt1[i] = pos[0];
      nxtd[i] = pos[dst[i] + 1];
      pos[dst[i]] = i;
    }
    f[st] = st;
    for (int i = st; i < n; i++) {
      if (nxt1[i] != n + 1) (f[nxt1[i]] += f[i]) %= mod;
      if (nxtd[i] != n + 1) (f[nxtd[i]] += f[i]) %= mod;
    }
    int ans = 0;
    for (int i = st; i <= n; i++)
      if (dst[i] <= dst[n]) (ans += f[i]) %= mod;
    printf("%d\n", ans);
  }
  return 0;
}
