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;
char buf[1000010];
string s;
int sum[1000100];
int len[1000100];
int low[2000100], high[2000100];
int main(void) {
  int i;
  scanf("%s", buf);
  s = buf;
  int N = s.length();
  sum[0] = N + 10;
  for ((i) = 0; (i) < (int)(N); (i)++)
    sum[i + 1] = sum[i] + ((s[i] == '(') ? 1 : -1);
  for ((i) = 0; (i) < (int)(2000100); (i)++) {
    low[i] = (1 << 29);
    high[i] = -(1 << 29);
  }
  for (i = N; i >= 0; i--) {
    if (i != N) {
      if (low[sum[i] - 1] != (1 << 29))
        len[i] = low[sum[i] - 1] - i - 1;
      else if (high[sum[i]] != -(1 << 29))
        len[i] = high[sum[i]] - i;
    }
    low[sum[i]] = min(low[sum[i]], i);
    high[sum[i]] = max(high[sum[i]], i);
  }
  int big = 0;
  for ((i) = 0; (i) < (int)(N); (i)++) big = max(big, len[i]);
  int cnt = 0;
  for ((i) = 0; (i) < (int)(N); (i)++)
    if (len[i] == big) cnt++;
  if (big == 0) {
    cout << 0 << ' ' << 1 << endl;
  } else {
    cout << big << ' ' << cnt << endl;
  }
  return 0;
}