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;
string s;
int cnt[1001000];
int id[1001000], nt;
int le[1001000];
void pop(int now) {
  if (nt == 0) return;
  nt--;
  if (id[nt] >= 1 && s[id[nt] - 1] == ')') {
    le[now] = le[id[nt] - 1];
  }
  le[now] += now - id[nt] + 1;
  cnt[le[now]]++;
}
void push(int now) {
  id[nt] = now;
  nt++;
}
int main() {
  cin >> s;
  int n = s.length();
  int le = 0, ri = 0;
  int ccc = 0;
  for (int i = (0); (i) <= (n - 1); (i)++) {
    if (s[i] == '(')
      push(i);
    else
      pop(i);
  }
  int maxlen = 0, maxn = 1;
  for (int i = (1); (i) <= (n); (i)++) {
    if (cnt[i]) maxlen = i, maxn = cnt[i];
  }
  cout << maxlen << " " << maxn << endl;
  return 0;
}