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 long long MOD = 1e9 + 7;
const int N = 1e6 + 5;
int n, nxt[N];
string s;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> s;
  n = s.length();
  stack<int> st;
  for (int i = 0; i < n; i++) {
    if (s[i] == '(') {
      st.push(i);
    } else {
      if (!st.empty()) {
        int e = st.top();
        st.pop();
        nxt[e] = i;
      }
    }
  }
  int ans = 0, cur = 0, cnt = 1;
  for (int i = 0; i < n; i++) {
    if (!nxt[i]) {
      cur = 0;
      continue;
    }
    cur += nxt[i] - i + 1;
    i += (nxt[i] - i);
    if (cur > ans) {
      ans = cur;
      cnt = 1;
    } else if (cur == ans)
      cnt++;
  }
  cout << ans << " " << cnt;
  return 0;
}