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;
long long int gcd(long long int a, long long int b) {
  return b ? gcd(b, a % b) : a;
}
long long int powmod(long long int a, long long int b) {
  long long int res = 1;
  a %= 998244253LL;
  assert(b >= 0);
  for (; b; b >>= 1) {
    if (b & 1) res = res * a % 998244253LL;
    a = a * a % 998244253LL;
  }
  return res;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  string second;
  cin >> second;
  int i, n = second.size(), mx = 0, frq = 0;
  int v[n];
  stack<int> stk;
  for (i = 0; i < n; i++) {
    if (second[i] == '(') {
      v[i] = 0;
      stk.push(i);
    } else {
      if (stk.empty())
        v[i] = 0;
      else {
        v[i] = i - stk.top() + 1;
        if (stk.top() > 0) v[i] += v[stk.top() - 1];
        stk.pop();
      }
    }
    if (v[i] > mx)
      mx = v[i], frq = 1;
    else if (v[i] == mx)
      frq++;
  }
  if (!mx)
    cout << "0 1";
  else
    cout << mx << ' ' << frq;
  return 0;
}