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 mod = 1e9 + 7;
int n, m, i, j, dp[1000005], sum, lst, mx;
char ss[1000005];
string st;
void upd(int x) {
  int tmp = sum, i;
  for (i = 0; i <= x; i++) {
    sum = (sum - dp[i] + mod) % mod;
    dp[i] = tmp;
    sum = (sum + dp[i]) % mod;
  }
}
int main() {
  scanf("%s", ss);
  st = ss;
  int c0 = 0, c1 = 0;
  while (!st.empty() && st[st.size() - 1] == '0') {
    c0++;
    st.erase(st.size() - 1);
  }
  if (st.empty()) {
    printf("%d\n", c0);
    return 0;
  } else {
    st.erase(st.size() - 1);
  }
  reverse(st.begin(), st.end());
  while (!st.empty() && st[st.size() - 1] == '0') {
    c1++;
    st.erase(st.size() - 1);
  }
  if (st.empty()) {
    printf("%d\n", 1ll * (c0 + 1) * (c1 + 1) % mod);
    return 0;
  } else {
    st.erase(st.size() - 1);
  }
  reverse(st.begin(), st.end());
  sum = 1;
  lst = 0;
  for (i = 0; i < st.size(); i++) {
    if (st[i] == '1') {
      mx = max(mx, lst);
      upd(lst);
      lst = 0;
    } else {
      lst++;
    }
  }
  upd(lst);
  printf("%d\n", 1ll * (c0 + 1) * (c1 + 1) % mod * sum % mod);
  return 0;
}
