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 long long const1 = 1e9 + 7;
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  string s;
  cin >> s;
  long long n = s.size();
  vector<long long> dist(n);
  for (long long i = 0; i < n; i++) {
    if (s[i] == '0') {
      if (i == 0) {
        dist[i] = 1;
      } else {
        dist[i] = dist[i - 1] + 1;
      }
    }
  }
  vector<long long> dp(n + 2), nxt(n + 2, n);
  for (long long i = n - 1; i >= 0; i--) {
    long long a, b;
    if (nxt[0] < n) {
      a = dp[nxt[0]];
    } else {
      a = 0;
    }
    if (dist[i] + 1 < n) {
      b = dp[nxt[dist[i] + 1]];
    } else {
      b = 0;
    }
    dp[i] = (a + b);
    if (dist[i] <= dist.back()) {
      dp[i]++;
    }
    dp[i] %= const1;
    nxt[dist[i]] = i;
  }
  long long ans = n;
  if (nxt[0] < n) {
    ans = dp[nxt[0]] * (nxt[0] + 1) % const1;
  }
  cout << ans;
  return 0;
}