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 mil = 1200000;
int st[3 * mil];
int x, y;
int main() {
  for (int i = 0; i < 3 * mil; i++) st[i] = -mil;
  int res = 0, cnt = 1, top = 0 + mil, cur = 0;
  int st_top = 0;
  string s;
  cin >> s;
  int len = s.length();
  st[top] = 0;
  for (int i = 0; i < len; i++) {
    if (s[i] == '(') {
      cur++;
      top++;
      st[top] = i + 1;
    }
    if (s[i] == ')') {
      cur--;
      top--;
      if (st[top] == -mil) st[top] = i + 1;
      int curres = i + 1 - st[top];
      if (curres > res) {
        res = curres;
        cnt = 1;
        x = st[top];
        y = i + 1;
      } else if (curres == res)
        cnt++;
      else
        ;
    }
  }
  if (res == 0) cnt = 1;
  cout << res << " " << cnt;
  return 0;
}