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;
template <class T>
inline void read(T &x) {
  x = 0;
  char c = getchar();
  int f = 1;
  while (!isdigit(c)) {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (isdigit(c)) {
    x = x * 10 + c - '0';
    c = getchar();
  }
  x *= f;
}
template <class T>
T gcd(T a, T b) {
  return !b ? a : gcd(b, a % b);
}
template <class T>
inline void umin(T &x, T y) {
  x = x < y ? x : y;
}
template <class T>
inline void umax(T &x, T y) {
  x = x > y ? x : y;
}
mt19937 R(chrono::system_clock().now().time_since_epoch().count());
const int N = 1e6 + 11, mo = 1e9 + 7;
inline void add(int &x, int y) { x = x + y < mo ? x + y : x + y - mo; }
inline void sub(int &x, int y) { x = x - y >= 0 ? x - y : x - y + mo; }
char s[N];
int res, n, f[2][N], a[N], len, nxt[N], sta[N], top;
int main() {
  scanf("%s", s + 1);
  n = strlen(s + 1);
  s[n + 1] = '1';
  for (register int i = (1); i <= (n); i++)
    if (s[i] == '1') {
      int j = i + 1;
      while (s[j] != '1') j++;
      a[++len] = j - i - 1;
    }
  int ways = 1;
  if (s[1] == '0') {
    while (ways <= n && s[ways] == '0') ways++;
    if (ways > n) {
      printf("%d\n", n);
      return 0;
    }
  }
  f[1][1] = ways;
  int lst = 0;
  while (s[n - lst] == '0') lst++;
  for (register int i = (len); i >= (1); i--) {
    while (top && a[sta[top]] <= a[i]) top--;
    if (top)
      nxt[i] = sta[top];
    else
      nxt[i] = len + 1;
    sta[++top] = i;
  }
  for (register int i = (1); i <= (len); i++) {
    int alr = 0;
    for (int j = i; j <= len;) {
      if (alr + 1 <= min(a[j], lst))
        add(res, 1ll * f[1][i] * (min(a[j], lst) - alr) % mo);
      add(f[0][j], 1ll * f[1][i] * (a[j] - alr) % mo);
      alr = a[j];
      j = nxt[j];
    }
    add(f[1][i + 1], f[0][i]);
    add(f[1][i + 1], f[1][i]);
  }
  for (register int i = (1); i <= (len); i++) {
    add(res, f[1][i]);
  }
  cout << res;
  return 0;
}
