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 MAXN = 1e6 + 5;
const int P = 1e9 + 7;
template <typename T>
void chkmax(T &x, T y) {
  x = max(x, y);
}
template <typename T>
void chkmin(T &x, T y) {
  x = min(x, y);
}
template <typename T>
void read(T &x) {
  x = 0;
  int f = 1;
  char c = getchar();
  for (; !isdigit(c); c = getchar())
    if (c == '-') f = -f;
  for (; isdigit(c); c = getchar()) x = x * 10 + c - '0';
  x *= f;
}
void update(int &x, int y) {
  x += y;
  if (x >= P) x -= P;
}
char s[MAXN];
int m, a[MAXN];
int n, sum, dp[MAXN], aux[MAXN];
int main() {
  scanf("%s", s + 1);
  n = strlen(s + 1);
  int last = 0;
  for (int i = 1; i <= n; i++)
    if (s[i] == '1') {
      a[++m] = i - last - 1;
      last = i;
    }
  if (last == 0) {
    cout << n << endl;
    return 0;
  }
  for (int i = m; i >= 1; i--) {
    dp[i] = (sum + n - last + 1) % P;
    for (int j = 0; j <= a[i]; j++) {
      update(sum, P - aux[j]);
      update(sum, dp[i]);
      aux[j] = dp[i];
    }
  }
  int ans = 0;
  for (int i = 0; s[i] != '1'; i++) update(ans, aux[i]);
  cout << ans << endl;
  return 0;
}
