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 N = 1e6 + 10;
int a[N], ct[N];
signed main() {
  string s;
  cin >> s;
  int n = s.length();
  fill(a, a + N, -1);
  stack<int> stk;
  for (int i = 0; i < n; i++) {
    if (s[i] == '(')
      stk.push(i);
    else {
      if (!stk.empty()) {
        int top = stk.top();
        a[i] = top;
        stk.pop();
        if (top > 0 and a[top - 1] != -1) a[i] = a[top - 1];
      }
    }
  }
  int mx = 0;
  for (int i = 0; i < n; i++)
    if (a[i] != -1) {
      ct[i - a[i] + 1]++;
      mx = max(mx, i - a[i] + 1);
    }
  ct[0] = 1;
  cout << mx << " " << ct[mx] << "\n";
}