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;
int n;
long long dp[1000005], last1, t, v, mod = 1e9 + 7;
string s;
struct nibab {
  long long s, l, v;
};
stack<nibab> q;
int32_t main() {
  ios_base ::sync_with_stdio(0);
  cin.tie();
  cout.tie();
  cin >> s;
  n = s.length();
  for (int i = 1; i <= n; i++) {
    if (s[i - 1] == '0') {
      t++;
      if (last1)
        dp[i] = dp[last1];
      else
        dp[i] = 1;
      if (i > 1 && s[i - 2] == '0') dp[i] = (dp[i] + dp[i - 1]) % mod;
      while (q.size() && t >= q.top().s) v -= q.top().v, q.pop();
      if (q.size()) {
        v -= q.top().v;
        q.top().v = dp[q.top().l - t];
        v += q.top().v;
      }
      v += dp[i];
      v = (v % mod + mod) % mod;
      q.push({t, i, dp[i]});
    } else {
      if (!last1 && q.size()) q.pop();
      t = 0;
      dp[i] = (dp[last1] + v + 1) % mod;
      last1 = i;
    }
  }
  if (s[n - 1] == '0')
    cout << (dp[last1] + dp[n]) % mod << '\n';
  else
    cout << dp[last1] << '\n';
}