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 = 1000005;
char s[maxn], ss[maxn];
pair<int, int> work(char s[maxn]) {
  int cnt = 0, ret = 0;
  int sum = 0;
  int start = 0;
  int lens = strlen(s);
  for (int i = 0; i < lens; i++) {
    if (s[i] == '(')
      sum++;
    else
      sum--;
    if (sum < 0)
      start = i + 1, sum = 0;
    else if (!sum) {
      if (i - start + 1 > ret)
        ret = i - start + 1, cnt = 1;
      else if (i - start + 1 == ret)
        cnt++;
    }
  }
  return make_pair(ret, cnt);
}
int main() {
  scanf("%s", s);
  int ans = 0, cnt;
  pair<int, int> tmp = work(s);
  if (tmp.first > ans) {
    ans = tmp.first;
    cnt = tmp.second;
  }
  int lens = strlen(s);
  for (int i = 0; i < lens; i++) ss[i] = s[lens - i - 1] == '(' ? ')' : '(';
  tmp = work(ss);
  if (tmp.first > ans) {
    ans = tmp.first;
    cnt = tmp.second;
  }
  if (ans)
    printf("%d %d", ans, cnt);
  else
    printf("0 1");
  return 0;
}