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 = 1000 * 1000 + 10;
int n, dp[N], mx, cnt;
string s;
int main() {
  ios::sync_with_stdio(false), cin.tie(NULL), cout.tie(NULL);
  fill(dp, dp + N, -1);
  cin >> s, n = s.size();
  for (int i = 1; i < n; i++) {
    if (s[i - 1] == '(' && s[i] == ')') dp[i] = i - 1;
    if (dp[i - 1] ^ -1 && s[dp[i - 1] - 1] == '(' && s[i] == ')')
      dp[i] = dp[i - 1] - 1;
    if (dp[i] > 0 && dp[dp[i] - 1] ^ -1) dp[i] = dp[dp[i] - 1];
  }
  for (int i = 0; i < n; i++) {
    if (i - dp[i] + 1 > mx && dp[i] ^ -1)
      mx = i - dp[i] + 1, cnt = 1;
    else if (i - dp[i] + 1 == mx && dp[i] ^ -1)
      cnt++;
  }
  if (!mx)
    cout << "0 1\n";
  else
    cout << mx << ' ' << cnt << '\n';
}