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;
int main() {
  string st;
  cin >> st;
  int open = 0, close = 0;
  int end = 0;
  stack<int> sta;
  vector<int> vt2(1000000 + 1, -1);
  vector<int> dp(1000000 + 1, -1);
  vector<int> freq(1000000 + 1, 0);
  int n = st.length();
  for (int i = 0; i < n; i++) {
    if (st[i] == ('(')) sta.push(i);
    if (st[i] == (')')) {
      if (!sta.empty()) {
        dp[i] = sta.top();
        vt2[i] = dp[i];
        sta.pop();
        if (dp[i] >= 1 and st[dp[i] - 1] == ')' and vt2[dp[i] - 1] >= 0) {
          vt2[i] = vt2[dp[i] - 1];
        }
        freq[i - vt2[i] + 1]++;
      }
    }
  }
  int ansflag = 0;
  for (int i = 0; i < st.length(); i++) {
  }
  for (int i = st.length(); i > 0; i--) {
    if (freq[i] > 0) {
      ansflag = 1;
      cout << i << " " << freq[i] << endl;
      break;
    }
  }
  if (ansflag == 0) cout << 0 << " " << 1 << endl;
}