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 + 16;
const int MO = 1e9 + 7;
int n, f[N], g[N], t[N], r;
char s[N];
void go(int x, int y) {
  r -= t[x];
  if (r < 0) r += MO;
  r += y;
  if (r >= MO) r -= MO;
  t[x] = y;
}
int main() {
  int i, o, x, y, z;
  scanf("%s", s + 1);
  n = strlen(s + 1);
  z = 1;
  while (s[z] == '0') {
    z++;
  }
  if (z > n) {
    printf("%d", n);
    return 0;
  }
  o = 0;
  for (i = z; i <= n; i = i + 1) {
    if (s[i] == '0') {
      o++;
      f[i] = (f[i - 1] + x) % MO;
      g[i] = g[i - 1];
      go(o, x);
    } else {
      o = 0;
      g[i] = (g[i - 1] + r + 1) % MO;
      f[i] = g[i];
      x = f[i];
    }
  }
  printf("%d", (long long)f[n] * z % MO);
  return 0;
}
