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 mm = 1001000;
char s[mm], q[mm];
int p[mm], pos, ans, num;
int main() {
  while (cin >> s) {
    int len = strlen(s);
    pos = 0;
    ans = 0;
    num = 0;
    p[0] = -1;
    for (int i = 0; i < len; i++) {
      if (!pos || s[i] == '(' || s[i] == q[pos]) {
        q[++pos] = s[i];
        p[pos] = i;
      } else
        pos--;
    }
    p[++pos] = len;
    for (int i = pos - 1; i >= 0; i--)
      if (p[i + 1] - p[i] - 1 > ans) {
        ans = p[i + 1] - p[i] - 1;
        num = 1;
      } else if (p[i + 1] - p[i] - 1 == ans)
        num++;
    if (ans == 0) num = 1;
    cout << ans << " " << num << "\n";
  }
}