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[1000010];
int d[1000010], vt[1000010];
int len;
int kqdodai, kqsl;
void timkq() {
  int i;
  int dodai;
  dodai = 0;
  kqdodai = 0;
  kqsl = 1;
  for (i = 0; i < len; i++) {
    if (d[i] == 1) {
      dodai++;
    }
    if (d[i] == 0 || i == len - 1) {
      if (dodai == kqdodai && dodai != 0) {
        kqsl++;
      }
      if (dodai > kqdodai) {
        kqdodai = dodai;
        kqsl = 1;
      }
      dodai = 0;
    }
  }
}
void solve() {
  int i;
  int tang, dodai;
  tang = 0;
  dodai = 0;
  len = strlen(s);
  for (i = 0; i < len; i++) {
    if (s[i] == '(') {
      vt[tang] = i;
      tang++;
    }
    if (s[i] == ')') {
      tang--;
      if (tang == -1) {
        tang = 0;
        continue;
      }
      d[vt[tang]] = 1;
      d[i] = 1;
    }
  }
  timkq();
}
void output() { printf("%d %d\n", kqdodai, kqsl); }
int main() {
  while (scanf("%s", &s) > 0) {
    solve();
    output();
  }
  return 0;
}