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;
int n;
char a[1000005];
int f[1000005] = {0};
int re[2] = {0, 1};
stack<int> s;
int main() {
  scanf("%s", &a);
  n = strlen(a);
  for (int i = 0; i < n; i++) {
    if (a[i] == '(')
      s.push(i);
    else {
      if (!s.empty()) {
        int t = s.top();
        s.pop();
        f[i] = f[t - 1] + i - t + 1;
      }
      if (f[i] > re[0]) {
        re[0] = f[i];
        re[1] = 1;
      } else if (f[i] == re[0]) {
        re[1]++;
      }
    }
  }
  if (re[0]) {
    printf("%d %d", re[0], re[1]);
  } else {
    printf("0 1", re[0], re[1]);
  }
  return 0;
}