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 s[1010101];
int pre[1010101];
int main() {
  while (scanf("%s", s) == 1) {
    memset(pre, -1, sizeof(pre));
    int i = 0;
    int mx = 0, cnt = 1;
    for (i = 0; s[i]; ++i) {
      if (i == 0 || s[i] == '(') continue;
      int j = i - 1;
      while (j >= 0 && s[j] == ')') j = pre[j] - 1;
      if (j < 0)
        continue;
      else if (s[j] == '(') {
        if (j - 1 >= 0 && pre[j - 1] >= 0) j = pre[j - 1];
        pre[i] = j;
        if (mx < i - j + 1) {
          mx = i - j + 1;
          cnt = 1;
        } else if (mx == i - j + 1) {
          cnt++;
        }
      }
    }
    printf("%d %d\n", mx, cnt);
  }
  return 0;
}