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 = 1000006;
char s[MAXN];
struct node {
  int num, ki;
} stk[MAXN];
bool f[MAXN];
int top, l, ans, ansk = 1, cnt;
int main() {
  scanf("%s", s + 1);
  l = strlen(s + 1);
  for (int i = 1; i <= l; ++i) {
    if (s[i] == '(') {
      stk[++top].num = 1;
      stk[top].ki = i;
    }
    if (s[i] == ')') {
      if (stk[top].num == 1) {
        f[stk[top].ki] = true;
        f[i] = true;
        top--;
      } else {
        stk[++top].num = 2;
        stk[top].ki = i;
      }
    }
  }
  for (int i = 1; i <= l; ++i) {
    if (f[i]) {
      ++cnt;
    } else {
      if (ans < cnt) {
        ans = cnt;
        ansk = 1;
      } else if (ans == cnt)
        ++ansk;
      cnt = 0;
    }
  }
  if (ans < cnt) {
    ans = cnt;
    ansk = 1;
  } else if (ans == cnt)
    ++ansk;
  ansk = ans ? ansk : 1;
  printf("%d %d\n", ans, ansk);
  return 0;
}