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;
stack<int> st;
char s[1000100];
int d[1000100];
int c[1000100];
int n;
void init(void) {
  scanf("%s", &s[1]);
  n = strlen(&s[1]);
  while (!st.empty()) st.pop();
}
void process(void) {
  int i, len, cnt;
  for (i = 1; i <= n; i = i + 1) {
    if (s[i] == '(')
      st.push(i);
    else {
      if (st.empty()) {
        d[i] = -1;
        c[i] = -1;
      } else {
        d[i] = st.top();
        st.pop();
        if (s[d[i] - 1] == ')' && c[d[i] - 1] > -1)
          c[i] = c[d[i] - 1];
        else
          c[i] = d[i];
      }
    }
  }
  len = 0;
  cnt = 0;
  for (i = 1; i <= n; i = i + 1)
    if (s[i] == ')' && c[i] > -1) {
      if (i - c[i] + 1 > len) {
        len = i - c[i] + 1;
        cnt = 1;
      } else if (i - c[i] + 1 == len)
        cnt++;
    }
  if (len == 0)
    printf("0 1");
  else
    printf("%d %d", len, cnt);
}
int main(void) {
  init();
  process();
  return 0;
}