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 inf = 1e6 + 9;
string a;
int n, pre[inf], smaller_left[inf];
stack<int> st;
map<int, vector<int> > mp;
int mx, cnt;
int main() {
  cin >> a;
  n = a.size();
  for (int i = 1; i <= n; i++) pre[i] = pre[i - 1] + (a[i - 1] == ')' ? -1 : 1);
  for (int i = 1; i <= n; i++) {
    while (!st.empty() && pre[st.top()] >= pre[i]) st.pop();
    smaller_left[i] = (st.empty() ? -1 : st.top());
    st.push(i);
  }
  mp[0].push_back(0);
  for (int i = 1; i <= n; i++) {
    int tmp =
        upper_bound(mp[pre[i]].begin(), mp[pre[i]].end(), smaller_left[i]) -
        mp[pre[i]].begin();
    if (tmp != mp[pre[i]].size()) {
      if (i - mp[pre[i]][tmp] > mx) mx = i - mp[pre[i]][tmp], cnt = 0;
      if (i - mp[pre[i]][tmp] == mx) cnt++;
    }
    mp[pre[i]].push_back(i);
  }
  if (mx == 0) cout << "0 1\n", exit(0);
  cout << mx << " " << cnt << endl;
}