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;
long long read() {
  long long x = 0;
  char f = 1, ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  return x * f;
}
void write(long long x) {
  if (x < 0) x = -x, putchar('-');
  if (x > 9) write(x / 10);
  putchar(x % 10 + '0');
}
inline void writeln(long long x) {
  write(x);
  puts("");
}
const int N = 5500000;
int st[N], top, n, cnt[N], f[N];
char s[N];
int main() {
  cin >> (s + 1);
  n = strlen(s + 1);
  for (int i = 1; i <= n; ++i)
    if (s[i] == '(')
      st[++top] = i;
    else if (top)
      ++cnt[f[i] = i - st[top] + 1 + f[st[top] - 1]], --top;
  cnt[0] = 1;
  for (int i = n; i >= 0; --i)
    if (cnt[i]) {
      printf("%d %d\n", i, cnt[i]);
      return 0;
    }
  return 0;
}