Problem: 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.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
template <class D>
inline void read(D &res) {
  res = 0;
  register int g = 1;
  register char ch = getchar();
  while (!isdigit(ch)) {
    if (ch == '-') g = -1;
    ch = getchar();
  }
  while (isdigit(ch)) {
    res = (res << 3) + (res << 1) + (ch ^ 48);
    ch = getchar();
  }
  res *= g;
}
const int Mod = 1e9 + 7;
char c[1010000];
int n, m, ans, a[1010000], dis[1010000], f[1010000], nt[1010000];
inline void add(int &x, int y) { (x += y) >= Mod ? x -= Mod : 0; }
int main() {
  cin >> c;
  n = strlen(c);
  for (register int i = 0; i <= n - 1; i++) a[i] = c[i] - '0';
  dis[0] = a[0] ^ 1;
  for (register int i = 1; i <= n - 1; i++)
    if (!a[i]) dis[i] = dis[i - 1] + 1;
  for (register int i = 0; i <= n - 1; i++) nt[i] = n;
  for (register int i = n - 1; i >= 0; i--) {
    if (dis[i] <= dis[n - 1])
      f[i] = 1;
    else
      f[i] = 0;
    if (nt[0] < n) add(f[i], f[nt[0]]);
    if (nt[dis[i] + 1] < n) add(f[i], f[nt[dis[i] + 1]]);
    nt[dis[i]] = i;
  }
  if (nt[0] < n)
    ans = (long long)f[nt[0]] * (nt[0] + 1) % Mod;
  else
    ans = n;
  cout << ans;
  return 0;
}