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 = 1e9 + 7;
const long long inf = 1e18;
const int maxn = 1e6 + 10;
const double eps = 1e-5;
int m, n, len;
char s[maxn];
int vis[maxn];
int main() {
  while (~scanf("%s", s)) {
    memset(vis, 0, sizeof(vis));
    int mx = -1;
    len = strlen(s);
    int cnt = 0;
    for (int i = 0; i < len; ++i) {
      if (s[i] == '(')
        cnt++;
      else if (s[i] == ')') {
        if (cnt) vis[i] = 1;
        cnt--;
      }
      if (cnt < 0) cnt = 0;
    }
    cnt = 0;
    for (int i = len - 1; i >= 0; --i) {
      if (s[i] == ')')
        cnt++;
      else if (s[i] == '(') {
        if (cnt) vis[i] = 1;
        cnt--;
      }
      if (cnt < 0) cnt = 0;
    }
    cnt = 0;
    n = 0;
    for (int i = 0; i < len; ++i) {
      if (vis[i])
        cnt++;
      else
        cnt = 0;
      if (cnt > mx) {
        mx = cnt;
        n = 1;
      } else if (cnt == mx)
        n++;
    }
    if (mx == 0) n = 1;
    printf("%d %d\n", mx, n);
  }
  return 0;
}