Problem: Filya just learned new geometry object — rectangle. He is given a field consisting of n × n unit cells. Rows are numbered from bottom to top with integer from 1 to n. Columns are numbered from left to right with integers from 1 to n. Cell, located at the intersection of the row r and column c is denoted as (r, c). Filya has painted two rectangles, such that their sides are parallel to coordinate axes and each cell lies fully inside or fully outside each of them. Moreover, no cell lies in both rectangles.

Later, hedgehog Filya became interested in the location of his rectangles but was unable to find the sheet of paper they were painted on. They were taken by Sonya and now she wants to play a little game with Filya. He tells her a query rectangle and she replies with the number of initial rectangles that lie fully inside the given query rectangle. The query rectangle should match the same conditions as initial rectangles. Rectangle lies fully inside the query if each o its cells lies inside the query.

Filya knows Sonya really well, so is sure that if he asks more than 200 questions she will stop to reply.

Input

The first line of the input contains an integer n (2 ≤ n ≤ 216) — size of the field.

For each query an integer between 0 and 2 is returned — the number of initial rectangles that lie fully inside the query rectangle.

Output

To make a query you have to print "? x1 y1 x2 y2" (without quotes) (1 ≤ x1 ≤ x2 ≤ n, 1 ≤ y1 ≤ y2 ≤ n), where (x1, y1) stands for the position of the bottom left cell of the query and (x2, y2) stands for the up right cell of the query. You are allowed to ask no more than 200 queries. After each query you should perform "flush" operation and read the answer.

In case you suppose you've already determined the location of two rectangles (or run out of queries) you should print "! x11 y11 x12 y12 x21 y21 x22 y22" (without quotes), where first four integers describe the bottom left and up right cells of the first rectangle, and following four describe the corresponding cells of the second rectangle. You can print the rectangles in an arbitrary order. After you have printed the answer, print the end of the line and perform "flush". Your program should terminate immediately after it print the answer.

Interaction

To flush you can use (just after printing an integer and end-of-line): 

  * fflush(stdout) in C++; 
  * System.out.flush() in Java; 
  * stdout.flush() in Python; 
  * flush(output) in Pascal; 
  * See the documentation for other languages. 



You will get the Wrong Answer verdict if you ask more than 200 queries, or if you print an incorrect coordinates.

You will get the Idleness Limit Exceeded verdict if you don't print anything (but you should) or if you forget about flushing the output (more info below).

Hacking.

The first line should contain an integer n (2 ≤ n ≤ 216).

The second line should contain four integers x1, y1, x2, y2 (1 ≤ x1 ≤ x2 ≤ n, 1 ≤ y1 ≤ y2 ≤ n) — the description of the first rectangle.

The third line contains the description of the second rectangle in the similar way.

Example

Input

5
2
1
0
1
1
1
0
1


Output

? 1 1 5 5
? 1 1 3 3
? 1 1 3 1
? 2 2 2 2
? 3 3 5 5
? 3 3 3 5
? 3 3 3 4
? 3 4 3 5
! 2 2 2 2 3 4 3 5

Solution (in JAVA):

Solution:
import java.io.*;
import java.util.*;

public class B {

	BufferedReader br;
	PrintWriter out;
	StringTokenizer st;
	boolean eof;
	
//	int X1, Y1, X2, Y2;
//	int X3, Y3, X4, Y4;
//	{
//		X1 = 1;
//		Y1 = 2;
//		X2 = 3;
//		Y2 = 4;
//		
//		X3 = 5;
//		Y3 = 1;
//		X4 = 5;
//		Y4 = 5;
//	}

	int ask(int x1, int y1, int x2, int y2) throws IOException {
		out.println("? " + x1 + " " + y1 + " " + x2 + " " + y2);
		out.flush();
		return nextInt();
	}
	
	int inside(int x1, int y1, int x2, int y2, int x3, int y3, int x4, int y4) {
		return inside(x1, x2, x3, x4) & inside(y1, y2, y3, y4);
	}
	
	int inside(int x1, int x2, int y1, int y2) {
		return (x1 <= y1 && y2 <= x2) ? 1 : 0;
	}
	
//	int ask(int x1, int y1, int x2, int y2) throws IOException {
//		return inside(x1, y1, x2, y2, X1, Y1, X2, Y2) + inside(x1, y1, x2, y2, X3, Y3, X4, Y4); 
//	}

	int askFlipped(int x1, int y1, int x2, int y2) throws IOException {
		return ask(y1, x1, y2, x2);
	}
	
	boolean check(int x1, int y1, int x2, int y2, int x3, int y3, int x4, int y4) throws IOException {
		if (x1 > x2 || y1 > y2 || x3 > x4 || y3 > y4) {
			return false;
		}
		
		if (Math.max(x1, x3) <= Math.min(x2, x4) && Math.max(y1, y3) <= Math.min(y2, y4)) {
			return false;
		}
		
		return check(x1, y1, x2, y2) && check(x3, y3, x4, y4);
	}
	
	boolean check(int x1, int y1, int x2, int y2) throws IOException {
		if (ask(x1, y1, x2, y2) != 1) {
			return false;
		}
		if (x1 != x2) {
			if (ask(x1 + 1, y1, x2, y2) != 0 || ask(x1, y1, x2 - 1, y2) != 0) {
				return false;
			}
		}
		
		if (y1 != y2) {
			if (ask(x1, y1 + 1, x2, y2) != 0 || ask(x1, y1, x2, y2 - 1) != 0) {
				return false;
			}
		}
		return true;
	}

	void solve() throws IOException {
		int n = nextInt();

		int low = 0; // 0
		int high = n; // >0
		while (high - low > 1) {
			int mid = (low + high) >> 1;
			int ret = ask(1, 1, mid, n);
			if (ret == 0) {
				low = mid;
			} else {
				high = mid;
			}
		}

		int minX2 = high;

		low = 0; // <2
		high = n; // 2
		while (high - low > 1) {
			int mid = (low + high) >> 1;
			int ret = ask(1, 1, mid, n);
			if (ret == 2) {
				high = mid;
			} else {
				low = mid;
			}
		}

		int maxX2 = high;

		low = 1; // >0
		high = n + 1; // 0
		while (high - low > 1) {
			int mid = (low + high) >> 1;
			int ret = ask(mid, 1, n, n);
			if (ret == 0) {
				high = mid;
			} else {
				low = mid;
			}
		}

		int maxX1 = low;

		low = 1; // 2
		high = n + 1; // <2
		while (high - low > 1) {
			int mid = (low + high) >> 1;
			int ret = ask(mid, 1, n, n);
			if (ret == 2) {
				low = mid;
			} else {
				high = mid;
			}
		}
		int minX1 = low;

		// / NOW Y

		low = 0; // 0
		high = n; // >0
		while (high - low > 1) {
			int mid = (low + high) >> 1;
			int ret = askFlipped(1, 1, mid, n);
			if (ret == 0) {
				low = mid;
			} else {
				high = mid;
			}
		}

		int minY2 = high;

		low = 0; // <2
		high = n; // 2
		while (high - low > 1) {
			int mid = (low + high) >> 1;
			int ret = askFlipped(1, 1, mid, n);
			if (ret == 2) {
				high = mid;
			} else {
				low = mid;
			}
		}

		int maxY2 = high;

		low = 1; // >0
		high = n + 1; // 0
		while (high - low > 1) {
			int mid = (low + high) >> 1;
			int ret = askFlipped(mid, 1, n, n);
			if (ret == 0) {
				high = mid;
			} else {
				low = mid;
			}
		}

		int maxY1 = low;

		low = 1; // 2
		high = n + 1; // <2
		while (high - low > 1) {
			int mid = (low + high) >> 1;
			int ret = askFlipped(mid, 1, n, n);
			if (ret == 2) {
				low = mid;
			} else {
				high = mid;
			}
		}
		int minY1 = low;

		int[] x1s = { minX1, maxX1 };
		int[] x2s = { minX2, maxX2 };
		int[] y1s = { minY1, maxY1 };
		int[] y2s = { minY2, maxY2 };
		
		for (int mask = 0; mask < 8; mask++) {
			int x1 = x1s[0];
			int x3 = x1s[1];
			
			int x2 = x2s[get(mask, 0)];
			int x4 = x2s[get(mask, 0) ^ 1];
			
			int y1 = y1s[get(mask, 1)];
			int y3 = y1s[get(mask, 1) ^ 1];
			
			int y2 = y2s[get(mask, 2)];
			int y4 = y2s[get(mask, 2) ^ 1];
			
			if (check(x1, y1, x2, y2, x3, y3, x4, y4)) {
				out.printf("! %d %d %d %d %d %d %d %d\n", x1, y1, x2, y2, x3, y3, x4, y4);
				out.flush();
				return;
			}
		}
	}
	
	int get(int mask, int i) {
		return (mask >> i) & 1;
	}

	B() throws IOException {
		br = new BufferedReader(new InputStreamReader(System.in));
		out = new PrintWriter(System.out);
		solve();
		out.close();
	}

	public static void main(String[] args) throws IOException {
		new B();
	}

	String nextToken() {
		while (st == null || !st.hasMoreTokens()) {
			try {
				st = new StringTokenizer(br.readLine());
			} catch (Exception e) {
				eof = true;
				return null;
			}
		}
		return st.nextToken();
	}

	String nextString() {
		try {
			return br.readLine();
		} catch (IOException e) {
			eof = true;
			return null;
		}
	}

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

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

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