Problem: This is an easier version of the problem. In this version, n ≤ 500.

Vasya is an experienced developer of programming competitions' problems. As all great minds at some time, Vasya faced a creative crisis. To improve the situation, Petya gifted him a string consisting of opening and closing brackets only. Petya believes, that the beauty of the bracket string is a number of its cyclical shifts, which form a correct bracket sequence.

To digress from his problems, Vasya decided to select two positions of the string (not necessarily distinct) and swap characters located at this positions with each other. Vasya will apply this operation exactly once. He is curious what is the maximum possible beauty he can achieve this way. Please help him.

We remind that bracket sequence s is called correct if: 

  * s is empty; 
  * s is equal to "(t)", where t is correct bracket sequence; 
  * s is equal to t_1 t_2, i.e. concatenation of t_1 and t_2, where t_1 and t_2 are correct bracket sequences. 



For example, "(()())", "()" are correct, while ")(" and "())" are not.

The cyclical shift of the string s of length n by k (0 ≤ k < n) is a string formed by a concatenation of the last k symbols of the string s with the first n - k symbols of string s. For example, the cyclical shift of string "(())()" by 2 equals "()(())".

Cyclical shifts i and j are considered different, if i ≠ j.

Input

The first line contains an integer n (1 ≤ n ≤ 500), the length of the string.

The second line contains a string, consisting of exactly n characters, where each of the characters is either "(" or ")".

Output

The first line should contain a single integer — the largest beauty of the string, which can be achieved by swapping some two characters.

The second line should contain integers l and r (1 ≤ l, r ≤ n) — the indices of two characters, which should be swapped in order to maximize the string's beauty.

In case there are several possible swaps, print any of them.

Examples

Input


10
()()())(()


Output


5
8 7


Input


12
)(()(()())()


Output


4
5 10


Input


6
)))(()


Output


0
1 1

Note

In the first example, we can swap 7-th and 8-th character, obtaining a string "()()()()()". The cyclical shifts by 0, 2, 4, 6, 8 of this string form a correct bracket sequence.

In the second example, after swapping 5-th and 10-th character, we obtain a string ")(())()()(()". The cyclical shifts by 11, 7, 5, 3 of this string form a correct bracket sequence.

In the third example, swap of any two brackets results in 0 cyclical shifts being correct bracket sequences. 

Solution (in JAVA):

Solution:

import java.io.*;
import java.util.*;

public class Main {
    static int inf = (int) 1e9 + 7;
    static int n, a[];

    static int min() {
        int ans = inf;
        for(int i = 0;i < n;i++) ans = Math.min(ans, a[i]);
        return ans;
    }

    static int cnt(int l, int r, int num) {
        int ans = 0;
        for(int i = l;i <= r;i++) if (a[i] == num) ans++;
        return ans;
    }

    static pair check(pair a) {
        if (a.l < 0 || a.r >= n) return new pair(0, 0, 0);
        return a;
    }

    static pair max(pair a, pair b) {
        a = check(a);
        b = check(b);
        if (a.ans > b.ans) return a;
        return b;
    }

    static void scan() throws IOException {
        n = sc.nextInt();
        char h[] = sc.next().toCharArray();
        a = new int [n];
        for(int i = 0;i < n;i++) {
            a[i] = (h[i] == '(' ? 1 : -1);
            if (i != 0) a[i] += a[i - 1];
        }
    }

    static pair zero() {
        return new pair(0, 0, cnt(0, n - 1, min()));
    }

    static pair plus() {
        int min = min();
        int l = inf;
        int r = -1;
        for(int i = 0;i < n;i++) {
            if (l == inf && a[i] == min) l = i;
            if (a[i] == min) r = i;
        }

        return max(plus1(l, r, min), plus2(l, r, min));
    }

    static pair plus1(int l, int r, int min) {
        int ans = cnt(0, l - 1, min + 1) + cnt(r + 1, n - 1, min + 1);
        return new pair(l, r + 1, ans);
    }

    static pair plus2(int l, int r, int min) {
        int l1 = l;
        int r1 = r;
        for(int i = 0;i < n;i++) {
            if (a[i] == min + 1 && l1 > i) l1 = i;
            if (a[i] == min + 1 && r1 < i) r1 = i;
        }


        int ans = cnt(0, n - 1, min);
        for(int i = 0;i < n;i++) {
            if ((i < l1 || i > r1) && a[i] == min + 2) ans++;
        }

        return new pair(l1, r1 + 1, ans);
    }

    static pair minus() {
        int min = min();
        int cnt = cnt(0, n - 1, min);
        int l = 0;
        pair ans = new pair(0, 0, 0);
        for(int i = 0;i <= n;i++) {
            if (i == n || a[i] == min) {
                ans = max(ans, minus(l, i - 1, min, cnt));
                l = i + 1;
            }
        }
        return ans;
    }

    static pair minus(int l, int r, int min, int cnt) {
        int cnt1 = 0;
        int cnt2 = 0;
        pair ans = new pair(0, 0, 0);
        int l1 = l;
        for(int i = l;i <= r;i++) {
            if (a[i] == min + 1) {
                cnt1++;
                ans = max(ans, minus2(l1, i - 1, min, cnt));
                l1 = i + 1;
            }

            if (a[i] == min + 2) cnt2++;

        }

        ans = max(ans, minus2(l1, r, min, cnt));

        if (cnt1 > 0) return max(new pair(l, r + 1, cnt1), ans);
        return new pair(l, r + 1, cnt + cnt2);
    }

    static pair minus2(int l, int r, int min, int cnt) {
        return new pair(l, r + 1, cnt(l, r, min + 2) + cnt);
    }

    public static void main(String[] args) throws IOException {
        sc = new Scanner(System.in);
        pw = new PrintWriter(System.out);
        scan();
        if (a[n - 1] != 0) {
            System.out.println(0);
            System.out.println(1 + " " + 1);
            return;
        }
        pair ans = max(zero(), max(plus(), minus()));
        pw.println(ans.ans);
        pw.println(ans.l + 1 + " " + (ans.r + 1));

        pw.close();
    }

    static Scanner sc;
    static PrintWriter pw;

    static class Scanner {
        BufferedReader br;
        StringTokenizer st = new StringTokenizer("");

        Scanner(InputStream in) throws FileNotFoundException {
            br = new BufferedReader(new InputStreamReader(in));
        }

        Scanner(String in) throws FileNotFoundException {
            br = new BufferedReader(new FileReader(in));
        }

        String next() throws IOException {
            while (!st.hasMoreTokens()) st = new StringTokenizer(br.readLine());
            return st.nextToken();
        }

        int nextInt() throws IOException {
            return Integer.parseInt(next());
        }

        long nextLong() throws IOException {
            return Long.parseLong(next());
        }

        double nextDouble() throws IOException {
            return Double.parseDouble(next());
        }
    }
}

class pair {
    int l, r, ans;

    pair(int l, int r, int ans) {
        this.l = l;
        this.r = r;
        this.ans = ans;
    }

    pair() {}
}