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 + 5;
char str[maxn];
int dp[maxn];
int main() {
  while (scanf("%s", &str) == 1) {
    int len = strlen(str);
    int ans = 0;
    int sum = 0;
    stack<int> s;
    memset(dp, 0, sizeof(dp));
    for (int i = 0; i < len; ++i) {
      if (str[i] == '(') {
        s.push(i);
      } else {
        if (!s.empty()) {
          int temp = s.top();
          s.pop();
          if (temp != 0) {
            dp[i] = dp[temp - 1] + i - temp + 1;
          } else {
            dp[i] = dp[0] + i - temp + 1;
          }
          if (ans < dp[i]) {
            ans = dp[i];
            sum = 1;
          } else if (ans == dp[i]) {
            ++sum;
          }
        }
      }
    }
    if (ans == 0) {
      printf("0 1\n");
    } else {
      printf("%d %d\n", ans, sum);
    }
  }
  return 0;
}