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>
const int N = 1e6 + 5, djq = 1e9 + 7;
int n, a[N], f[N], ans, nxt[N], top, stk[N], pre[N], sum[N], endc;
char s[N];
int main() {
  scanf("%s", s + 1);
  n = strlen(s + 1);
  int lst = n + 1;
  while (s[n - endc] == '0') endc++;
  if (endc == n) return std::cout << n << std::endl, 0;
  for (int i = 1; i <= n; i++) a[i] = s[i] == '1' ? 0 : a[i - 1] + 1;
  for (int i = n; i >= 0; i--)
    if (nxt[i] = lst, s[i] == '1') lst = i;
  f[0] = sum[0] = 1;
  stk[top = 0] = -1;
  for (int i = 1; i <= n; i++) {
    while (top && a[stk[top]] < a[i]) top--;
    pre[i] = stk[top];
    stk[++top] = i;
  }
  for (int i = 0; i <= n; i++) {
    if (i) sum[i] = sum[i - 1];
    if (s[i] == '0') {
      f[i] = sum[i];
      if (pre[i] >= 0)
        f[i] = (f[i] - sum[pre[i]] + djq) % djq;
      else if (a[i] < i)
        f[i] = (f[i] + djq - 1) % djq;
    } else if (i)
      sum[i] = (sum[i] + f[i]) % djq;
    f[nxt[i]] = (f[nxt[i]] + f[i]) % djq;
  }
  for (int i = 1; i <= n; i++)
    if (s[i] == '1') ans = (1ll * (endc + 1) * f[i] + ans) % djq;
  return std::cout << ans << std::endl, 0;
}
