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;
vector<long long int> v;
map<long long int, long long int> ww, u;
vector<pair<long long int, long long int> > uu;
vector<long long int> uuu[200005];
map<pair<long long int, long long int>, long long int> vo;
set<long long int> uo;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long int n, m, i, j, k, p, q, o, l, s, tt = 1, t, z, aa;
  string a, b, c;
  double x, y, d, e, f;
  while (tt--) {
    cin >> a;
    k = 0;
    z = 0;
    stack<long long int> re;
    re.push(-1);
    u[0] = 1;
    for (i = 0; a[i]; i++) {
      if (a[i] == '(') {
        re.push(i);
      } else {
        re.pop();
        if (re.empty()) {
          re.push(i);
        } else {
          k = i - re.top();
          u[k]++;
          z = max(z, k);
        }
      }
    }
    cout << z << " " << u[z] << "\n";
  }
  return 0;
}