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>
const double PI = 3.141592653589793238460;
using namespace std;
int dp[1000001], C[1000001], D[1000001];
stack<int> stk;
int main() {
  string st;
  int mx = 0;
  cin >> st;
  for (int i = 0; i < st.size(); i++) {
    char ch = st[i];
    if (ch == '(') {
      stk.push(i);
    } else {
      if (stk.empty()) {
        C[i] = D[i] = -1;
      } else {
        int pos = stk.top();
        stk.pop();
        C[i] = D[i] = pos;
        if (pos > 0 && st[pos - 1] == ')' && C[pos - 1] != -1)
          C[i] = C[pos - 1];
        dp[i - C[i] + 1]++;
        mx = max(mx, i - C[i] + 1);
      }
    }
  }
  dp[0] = 1;
  cout << mx << " " << dp[mx];
}