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;
const int N = 1e6 + 6, mod = 1e9 + 7;
int dp[N], tmp[N], lv[N];
char s[N];
int main() {
  scanf("%s", s);
  int n = strlen(s);
  for (int i = n - 1; i >= 0; --i) {
    s[i] -= '0';
  }
  if (s[0])
    tmp[0] = 1;
  else
    dp[0] = tmp[0] = 1;
  int cur = 0, one = 0;
  for (int i = 0; i < n; ++i) {
    if (s[i]) {
      cur = 0;
      ++one;
    } else {
      ++cur;
    }
    lv[i] = cur;
    dp[i] = (dp[i] + tmp[cur]) % mod;
    tmp[cur] = 0;
    if (s[i]) {
      tmp[0] = (tmp[0] + dp[i]) % mod;
      tmp[1] = (tmp[1] + dp[i]) % mod;
    } else {
      if (one) {
        tmp[0] = (tmp[0] + dp[i]) % mod;
        tmp[cur + 1] = (tmp[cur + 1] + dp[i]) % mod;
      } else {
        tmp[0] = (tmp[0] + dp[i]) % mod;
        if (s[i + 1] == 0) dp[i + 1] = (dp[i + 1] + dp[i]) % mod;
      }
    }
  }
  long long ans = 0;
  int biggest = 0;
  int ada = 0;
  for (int i = n - 1; i >= 0; --i) {
    if (s[i]) {
      ans = (ans + dp[i]) % mod;
      --one;
      ++ada;
    } else {
      if (ada) {
        if (one && lv[i] <= biggest) {
          ans = (ans + dp[i]) % mod;
        }
      } else {
        ans = (ans + dp[i]) % mod;
      }
      if (!ada) biggest = max(biggest, lv[i]);
    }
  }
  printf("%lld\n", ans);
  return 0;
}