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, md = 1e9 + 7;
char str[maxn];
map<int, int> cnt;
int main() {
  gets(str);
  int n = strlen(str), res = 0;
  vector<int> dp(n + 2, 0);
  stack<int> st;
  for (int i = 0; i <= n - 1; ++i) {
    if (st.empty() || (str[st.top()] == str[i]))
      st.push(i);
    else if (str[st.top()] == '(' && str[i] == ')') {
      int x = st.top();
      st.pop();
      res = max(res, i - x + 1 + (x - 1 >= 0 ? dp[x - 1] : 0));
      dp[i] = i - x + 1 + (x - 1 >= 0 ? dp[x - 1] : 0);
      cnt[dp[i]]++;
    } else {
      st.push(i);
    }
  }
  if (res)
    cout << res << " " << cnt[res] << endl;
  else
    puts("0 1");
}