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 MAXN = 1e6 + 10;
const int INF = 0x3f3f3f3f;
char s[MAXN];
int len[MAXN];
stack<int> S;
int main(void) {
  while (scanf("%s", s) == 1) {
    while (!S.empty()) S.pop();
    memset(len, 0, sizeof(len));
    int mx = 0;
    int cnt = 0;
    for (int i = 0; s[i]; ++i) {
      if (s[i] == '(')
        S.push(i);
      else {
        if (!S.empty()) {
          int p = S.top();
          S.pop();
          len[i] = (p > 0 ? len[p - 1] : 0) + (i - p + 1);
        }
      }
      if (len[i] > mx) {
        mx = len[i];
        cnt = 1;
      } else if (len[i] == mx) {
        cnt++;
      }
    }
    if (mx == 0)
      puts("0 1");
    else
      printf("%d %d\n", mx, cnt);
  }
  return 0;
}