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 p[1000005];
string s;
int main() {
  while (cin >> s) {
    memset(p, -1, sizeof(p));
    int i, j;
    int top = 0;
    int bottom = 0;
    int maxx = -1;
    int cnt = 0;
    for (i = 0; i < s.length(); i++) {
      if (s[i] == '(') {
        if (p[top] == -1) {
          p[top] = i;
        }
        top++;
      } else {
        if (top == bottom) {
          top++;
          bottom++;
        } else {
          if (i - (p[top - 1] - 1) > maxx) {
            maxx = i - (p[top - 1] - 1);
            cnt = 1;
          } else if (i - (p[top - 1] - 1) == maxx) {
            cnt++;
          }
          p[top--] = -1;
        }
      }
    }
    if (maxx == -1)
      cout << 0 << " " << 1 << endl;
    else
      cout << maxx << " " << cnt << endl;
  }
  return 0;
}