Problem: This is yet another problem dealing with regular bracket sequences.

We should remind you that a bracket sequence is called regular, if by inserting «+» and «1» into it we can get a correct mathematical expression. For example, sequences «(())()», «()» and «(()(()))» are regular, while «)(», «(()» and «(()))(» are not. 

You are given a string of «(» and «)» characters. You are to find its longest substring that is a regular bracket sequence. You are to find the number of such substrings as well.

Input

The first line of the input file contains a non-empty string, consisting of «(» and «)» characters. Its length does not exceed 106.

Output

Print the length of the longest substring that is a regular bracket sequence, and the number of such substrings. If there are no such substrings, write the only line containing "0 1".

Examples

Input

)((())))(()())


Output

6 2


Input

))(


Output

0 1

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int N = 2e5 + 15;
const long long int inf = 1e18;
void solve() {
  string s;
  cin >> s;
  long long int n = s.size();
  s = "a" + s;
  stack<long long int> st;
  vector<long long int> dp(n + 10);
  vector<long long int> fre(n + 10);
  for (long long int i = 1; i <= n; i++) {
    long long int cur = 0;
    if (s[i] == '(')
      st.push(i);
    else {
      if (!st.empty()) {
        cur = dp[st.top() - 1] + (i - st.top() + 1);
        fre[cur]++;
        st.pop();
      }
    }
    dp[i] = cur;
  }
  fre[0] = 1;
  for (long long int i = n; i >= 0; i--) {
    if (fre[i]) {
      cout << i << " " << fre[i] << "\n";
      return;
    }
  }
}
void debug(long long int tt) {}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int t = 1;
  while (t--) {
    solve();
  }
}