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.util.*;

//import tr1.FenwickTree;

import java.io.*;

public class tr0 {
	static PrintWriter out;
	static StringBuilder sb;
	static final double EPS = 1e-9;
	static long mod = 998244353;
	static int inf = (int) 1e9 + 2;
	static long[] fac;
	static int[] a;
	static int[] si;
	static ArrayList<Integer> primes;
	static ArrayList<Integer>[] ad;
	static ArrayList<pair>[] d;
	static edge[] ed;
	static int[][] vis;
	static int[] l, ch;
	static int[] occ, p;
	static Queue<Integer>[] can;
	static String s;
	static long[][] memo;
	static int n, m, k;
	static qu[] ar;
	static int[] d1, d2;

	public static void main(String[] args) throws Exception {
		Scanner sc = new Scanner(System.in);
		out = new PrintWriter(System.out);
		int n = sc.nextInt();
		char[] a = sc.nextLine().toCharArray();
		int ans = ti(a);
		int l = 1;
		int f = 1;
        // System.out.println(ans);
		for (int i = 0; i < n; i++) {
			for (int j = i + 1; j < n; j++) {
				if(a[i]==a[j])
					continue;
				char[] b = a.clone();
				b[i] = a[j];
				b[j] = a[i];
				int v = ti(b);
				if (v > ans) {
					ans = v;
					l = i+1;
					f = j+1;
				}
			}
		}
		out.println(ans);
		out.println(f + " " + l);
		out.flush();
	}

	static int ti(char[] a) {
		int ans = 0;
		int min = 0;
		int b = 0;
		for (int i = 0; i < a.length; i++) {
			if (a[i] == '(')
				b++;
			else
				b--;
			min = Math.min(min, b);
		}
		if (b != 0)
			return 0;
       for(int i=0;i<a.length;i++)
       {
    	   if(a[i]=='(')
    		   b++;
    	   else
    		   b--;
    	   if(b==min)
    		   ans++;
       }
		return ans;
	}

	static class qu implements Comparable<qu> {
		long a;
		int b, c;

		qu(long i, int v, int x) {
			a = i;
			b = v;
			c = x;
		}

		public String toString() {
			return a + " " + b + " " + c;
		}

		public int compareTo(qu o) {
			return 0;
		}
	}

	static class seg implements Comparable<seg> {
		int a;
		int b;

		seg(int s, int e) {
			a = s;
			b = e;
		}

		public String toString() {
			return a + " " + b;
		}

		public int compareTo(seg o) {
			// if(a==o.a)
			return o.b - b;
			// return
		}
	}

	static class pair {
		int to;
		int number;

		pair(int t, int n) {
			number = n;
			to = t;
		}

		public String toString() {
			return to + " " + number;
		}
	}

	static int modPow(int a, int e)

	{
		int res = 1;
		while (e > 0) {
			if ((e & 1) == 1)
				res = (res * a);
			a = (a * a);
			e >>= 1;
		}
		return res;
	}

	static long inver(long x) {
		int a = (int) x;
		long e = (mod - 2);
		long res = 1;
		while (e > 0) {
			if ((e & 1) == 1) {
				res = (int) ((1l * res * a) % mod);
			}
			a = (int) ((1l * a * a) % mod);
			e >>= 1;
		}
		return res % mod;
	}

	static class edge implements Comparable<edge> {
		int from;
		int to;
		int number;

		edge(int f, int t, int n) {
			from = f;
			to = t;
			number = n;
		}

		public String toString() {
			return from + " " + to + " " + number;
		}

		public int compareTo(edge f) {
			return f.number - number;
		}
	}

	static void seive(int N) {
		si = new int[N];
		primes = new ArrayList<>();
		si[1] = 1;
		for (int i = 2; i < N; i++) {
			if (si[i] == 0) {
				si[i] = i;
				primes.add(i);
			}
			for (int j = 0; j < primes.size() && primes.get(j) <= si[i] && (i * primes.get(j)) < N; j++)
				si[primes.get(j) * i] = primes.get(j);

		}
	}

	static long fac(int n) {
		if (n == 0)
			return fac[n] = 1;
		if (n == 1)
			return fac[n] = 1;
		long ans = 1;
		for (int i = 1; i <= n; i++)
			fac[i] = ans = (i % mod * ans % mod) % mod;
		return ans % mod;
	}

	static long gcd(long a, long b) {

		if (b == 0)
			return a;
		return gcd(b, a % b);
	}

	static class unionfind {
		int[] p;
		int[] size;
		int[] max;
		int num;

		unionfind(int n) {
			p = new int[n];
			size = new int[n];
			max = new int[n];
			for (int i = 0; i < n; i++) {
				p[i] = i;
				max[i] = i;
			}
			Arrays.fill(size, 1);
			num = n;
		}

		int findSet(int v) {
			if (v == p[v])
				return v;

			max[v] = Math.max(max[v], max[p[v]]);

			p[v] = findSet(p[v]);

			max[v] = Math.max(max[v], max[p[v]]);

			return p[v];
		}

		boolean sameSet(int a, int b) {
			a = findSet(a);
			b = findSet(b);
			if (a == b)
				return true;
			return false;
		}

		int max() {
			int max = 0;
			for (int i = 0; i < size.length; i++)
				if (size[i] > max)
					max = size[i];
			return max;
		}

		boolean combine(int a, int b) {
			a = findSet(a);
			b = findSet(b);
			if (a == b)
				return true;
			// System.out.println(num+" ppp");
			num--;
			if (size[a] > size[b]) {
				p[b] = a;
				max[a] = Math.max(max[a], max[b]);
				size[a] += size[b];

			} else {
				p[a] = b;
				max[b] = Math.max(max[a], max[b]);
				size[b] += size[a];
			}
			return false;
		}
	}

	static class Scanner {
		StringTokenizer st;
		BufferedReader br;

		public Scanner(InputStream system) {
			br = new BufferedReader(new InputStreamReader(system));
		}

		public Scanner(String file) throws Exception {
			br = new BufferedReader(new FileReader(file));
		}

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

		public String nextLine() throws IOException {
			return br.readLine();
		}

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

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

		public char nextChar() throws IOException {
			return next().charAt(0);
		}

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

		public boolean ready() throws IOException {
			return br.ready();
		}

		public void waitForInput() throws InterruptedException {
			Thread.sleep(3000);
		}
	}
}