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;
long long f[1000005], ans = 0;
int n, st, p[1000005], d[1000005], g[1000005], h[1000005];
char s[1000005];
void A(long long &x, long long y) {
  x += y;
  if (x > 1000000007) x -= 1000000007;
}
int main() {
  scanf("%s", s + 1);
  n = strlen(s + 1);
  p[0] = n + 1;
  for (register int i = 1; i <= n; i++) {
    p[i] = n + 1, d[i] = d[i - 1] + 1;
    if (s[i] == '1') {
      if (!st) st = i;
      d[i] = 0;
    }
  }
  if (!st) {
    printf("%d\n", n);
    return 0;
  }
  f[st] = st;
  for (register int i = n; i >= 1; i--) {
    g[i] = p[d[0]];
    h[i] = p[d[i] + 1];
    p[d[i]] = i;
  }
  for (register int i = st; i <= n; i++) {
    if (g[i] <= n) A(f[g[i]], f[i]);
    if (h[i] <= n) A(f[h[i]], f[i]);
  }
  for (register int i = 1; i <= n; i++)
    if (d[i] <= d[n]) A(ans, f[i]);
  printf("%lld\n", ans);
  return 0;
}
