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 maxn = 1e6 + 10;
stack<int> stk;
int m[maxn], dp[maxn];
int main() {
  string s;
  cin >> s;
  int n = s.size();
  for (int i = 0; i <= maxn; i++) m[i] = -1;
  for (int i = 0; i < n; i++) {
    if (s[i] == '(')
      stk.push(i);
    else if (stk.size() != 0) {
      m[i] = stk.top();
      m[stk.top()] = i;
      stk.pop();
    }
  }
  for (int i = 0; i < n; i++) {
    if (m[i] != -1 && s[i] == ')') {
      dp[i] = i - m[i] + 1;
      if (m[i] - 1 >= 0) dp[i] += dp[m[i] - 1];
    }
  }
  int maxx = 0, t = 0;
  for (int i = 0; i < n; i++) maxx = max(maxx, dp[i]);
  for (int i = 0; i < n; i++)
    if (dp[i] == maxx) t++;
  if (maxx == 0)
    cout << "0 1";
  else
    cout << maxx << " " << t;
  return 0;
}