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 mod = 1000000007;
struct custom_hash {
  static uint64_t splitmix64(uint64_t x) {
    x += 0x9e3779b97f4a7c15;
    x = (x ^ (x >> 30)) * 0xbf58476d1ce4e5b9;
    x = (x ^ (x >> 27)) * 0x94d049bb133111eb;
    return x ^ (x >> 31);
  }
  size_t operator()(uint64_t x) const {
    static const uint64_t FIXED_RANDOM =
        chrono::steady_clock::now().time_since_epoch().count();
    return splitmix64(x + FIXED_RANDOM);
  }
};
void solve() {
  int n, mx = 0;
  string st;
  unordered_map<int, int, custom_hash> S, E, F;
  F[0] = 1;
  cin >> st;
  n = ((int)(st).size());
  stack<int> stk;
  for (int i = 0; i < n; i++) {
    char c = st[i];
    if (c == '(') {
      stk.push(i);
    } else {
      if (stk.empty()) {
        S[i] = E[i] = -1;
      } else {
        int idx = stk.top();
        stk.pop();
        S[i] = E[i] = idx;
        if (idx > 0 && S[idx - 1] >= 0 && st[idx - 1] == ')') {
          E[i] = E[idx - 1];
        }
        int len = i - E[i] + 1;
        F[len]++;
        mx = max(mx, len);
      }
    }
  }
  cout << mx << ' ' << F[mx] << '\n';
  return;
}
int32_t main() {
  ios_base::sync_with_stdio(0), cout.tie(0), cin.tie(0);
  int t = 1;
  while (t--) {
    solve();
  }
}