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>
void read(T &x) {
  T f = 1;
  x = 0;
  char c = getchar();
  while (c < '0' || c > '9') {
    if (c == '-') f = -1;
    c = getchar();
  }
  while ('0' <= c && c <= '9') {
    x = x * 10 + c - '0';
    c = getchar();
  }
  x *= f;
}
template <class T>
void print(T x, char c) {
  if (x < 0) putchar('-'), x = -x;
  int s[20], top = 0;
  while (s[++top] = x % 10, x /= 10)
    ;
  while (top) putchar(s[top--] + '0');
  putchar(c);
}
int n, len[1000005], p, trans[1000005][2], lst[1000005], dp[1000005];
char s[1000005];
void add(int &x, int y) {
  x += y;
  if (x >= 1000000007) x -= 1000000007;
}
int main() {
  scanf("%s", s + 1);
  n = strlen(s + 1);
  for (int i = 1; i <= n; ++i)
    if (s[i] == '1') {
      p = i;
      break;
    }
  if (!p) return print(n, '\n'), 0;
  for (int i = 1; i <= n; ++i) {
    if (s[i] == '1')
      len[i] = 0;
    else
      len[i] = len[i - 1] + 1;
  }
  for (int i = n; i >= 1; --i) {
    trans[i][1] = lst[0];
    trans[i][0] = lst[len[i] + 1];
    lst[len[i]] = i;
  }
  dp[p] = p;
  int ans = 0;
  for (int i = p; i <= n; ++i) {
    if (trans[i][1]) add(dp[trans[i][1]], dp[i]);
    if (trans[i][0]) add(dp[trans[i][0]], dp[i]);
    if (len[i] <= len[n]) add(ans, dp[i]);
  }
  print(ans, '\n');
}
