Problem: This is an interactive problem.

This is a hard version of the problem. The difference from the easy version is that in the hard version 1 ≤ t ≤ min(n, 10^4) and the total number of queries is limited to 6 ⋅ 10^4.

Polycarp is playing a computer game. In this game, an array consisting of zeros and ones is hidden. Polycarp wins if he guesses the position of the k-th zero from the left t times.

Polycarp can make no more than 6 ⋅ 10^4 requests totally of the following type: 

  * ? l r — find out the sum of all elements in positions from l to r (1 ≤ l ≤ r ≤ n) inclusive. 



To make the game more interesting, each guessed zero turns into one and the game continues on the changed array. More formally, if the position of the k-th zero was x, then after Polycarp guesses this position, the x-th element of the array will be replaced from 0 to 1.

Help Polycarp win the game.

Interaction

First, your program must read two integers n and t (1 ≤ n ≤ 2 ⋅ 10^5, 1 ≤ t ≤ min(n, 10^4)).

Then t lines follow, each of which contains one integer k (1 ≤ k ≤ n). It is guaranteed that at the moment of the request the array contains at least k zeros. In order to get the next value of k, you must output the answer for the previous value of k.

After that, you can make no more than 6 ⋅ 10^4 requests in total.

Use the following format to output the answer (it is not a request, it doesn't count in 6 ⋅ 10^4): 

  * ! x — position of the k-th zero. 



Positions in the array are numbered from left to right from 1 to n inclusive.

After printing t answers, your program should exit immediately.

In this task, the interactor is not adaptive. This means that within the same test, the hidden array and the queries do not change.

In case of an incorrect query, -1 will be displayed. When this value is received, your program must immediately exit normally (for example, by calling exit(0)), otherwise, the testing system may issue an arbitrary verdict.

If the number of requests is exceeded, the verdict wrong answer will be displayed.

Your solution may get the verdict Idleness limit exceeded if you don't print anything or forget to flush the output buffer.

To flush the output buffer, you need to do the following immediately after the query output and the end-of-line character:

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



Hacks

Use the following format for hacks:

On the first line print the string s (1 ≤ |s| ≤ 2 ⋅ 10^5), consisting of zeros and ones, and an integer t (1 ≤ t ≤ min(|s|, 10^4)) — hidden array and number of requests, respectively. In the next t lines output the number k (1 ≤ k ≤ |s|).

The hacked solution will not have direct access to the hidden array.

Example

Input


6 2

2

2

1

1

0

1

0

Output


? 4 6

? 1 1

? 1 2

? 5 5

! 5

? 2 2

! 2

Note

In the first test, the array [1, 0, 1, 1, 0, 1] is hidden. After answering the query k=2, the array changed to [1, 0, 1, 1, 1, 1].

Solution (in JAVA):

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

/**
 *
 * @Har_Har_Mahadev
 */

/**
 * Main , Solution , Remove Public
 */
public class F {

	private static int[] arr;

	public static void process() throws IOException {

		int n = sc.nextInt();
		int t = sc.nextInt();
		arr = new int[n+1];
		Arrays.fill(arr, -1);
		while(t-- != 0) {
			int k = sc.nextInt();
			int l = 1,r = n;
			while(l < r) {
				int mid = (l+r)/2;
				int ff = -1;
				if(arr[mid] != -1)ff = arr[mid];
				else {
					ff = ask(1, mid);
					arr[mid] = ff;
				}
				 
				int cc = (mid-ff);
				if(cc<k)l = mid+1;
				else r = mid;
			}
			pflush("! "+r);
			for(int j = r; j<=n; j++) {
				if(arr[j] != -1)arr[j]++;
			}
		}
		
	}

	

	private static int ask(int i, int j) throws IOException {
		pflush("? "+i+" "+j);
		return sc.nextInt();
	}

	//=============================================================================
	//--------------------------The End---------------------------------
	//=============================================================================
	private static long INF = 2000000000000000000L, M = 1000000007, MM = 998244353;
	private static int N = 0;

	private static void google(int tt) {
		System.out.print("Case #" + (tt) + ": ");
	}

	static FastScanner sc;
	static PrintWriter out;

	public static void main(String[] args) throws IOException {
		boolean oj = true;
		if (oj) {
			sc = new FastScanner();
			out = new PrintWriter(System.out);
		} else {
			sc = new FastScanner(100);
			out = new PrintWriter("output.txt");
		}
		int t = 1;
//		t = sc.nextInt();
		int TTT = 1;
		while (t-- > 0) {
			//			google(TTT++);
			process();
		}
		out.flush();
		out.close();
	}

	static class Pair implements Comparable<Pair> {
		int x, y;

		Pair(int x, int y) {
			this.x = x;
			this.y = y;
		}

		@Override
		public int compareTo(Pair o) {
			return Integer.compare(this.x, o.x);
		}

		//		 @Override
		//		    public boolean equals(Object o) {
		//		        if (this == o) return true;
		//		        if (!(o instanceof Pair)) return false;
		//		        Pair key = (Pair) o;
		//		        return x == key.x && y == key.y;
		//		    }
		//		 
		//		    @Override
		//		    public int hashCode() {
		//		        int result = x;
		//		        result = 31 * result + y;
		//		        return result;
		//		    }
	}

	/////////////////////////////////////////////////////////////////////////////////////////////////////////

	static void println(Object o) {
		out.println(o);
	}

	static void println() {
		out.println();
	}

	static void print(Object o) {
		out.print(o);
	}

	static void pflush(Object o) {
		out.println(o);
		out.flush();
	}

	static int ceil(int x, int y) {
		return (x % y == 0 ? x / y : (x / y + 1));
	}

	static long ceil(long x, long y) {
		return (x % y == 0 ? x / y : (x / y + 1));
	}

	static int max(int x, int y) {
		return Math.max(x, y);
	}

	static int min(int x, int y) {
		return Math.min(x, y);
	}

	static int abs(int x) {
		return Math.abs(x);
	}

	static long abs(long x) {
		return Math.abs(x);
	}

	static long sqrt(long z) {
		long sqz = (long) Math.sqrt(z);
		while (sqz * 1L * sqz < z) {
			sqz++;
		}
		while (sqz * 1L * sqz > z) {
			sqz--;
		}
		return sqz;
	}

	static int log2(int N) {
		int result = (int) (Math.log(N) / Math.log(2));
		return result;
	}

	static long max(long x, long y) {
		return Math.max(x, y);
	}

	static long min(long x, long y) {
		return Math.min(x, y);
	}

	public static int gcd(int a, int b) {
		BigInteger b1 = BigInteger.valueOf(a);
		BigInteger b2 = BigInteger.valueOf(b);
		BigInteger gcd = b1.gcd(b2);
		return gcd.intValue();
	}

	public static long gcd(long a, long b) {
		BigInteger b1 = BigInteger.valueOf(a);
		BigInteger b2 = BigInteger.valueOf(b);
		BigInteger gcd = b1.gcd(b2);
		return gcd.longValue();
	}

	public static long lcm(long a, long b) {
		return (a * b) / gcd(a, b);
	}

	public static int lcm(int a, int b) {
		return (a * b) / gcd(a, b);
	}

	public static int lower_bound(int[] arr, int x) {
		int low = 0, high = arr.length, mid = -1;

		while (low < high) {
			mid = (low + high) / 2;

			if (arr[mid] >= x)
				high = mid;
			else
				low = mid + 1;
		}

		return low;
	}

	public static int upper_bound(int[] arr, int x) {
		int low = 0, high = arr.length, mid = -1;

		while (low < high) {
			mid = (low + high) / 2;

			if (arr[mid] > x)
				high = mid;
			else
				low = mid + 1;
		}

		return low;
	}

	/////////////////////////////////////////////////////////////////////////////////////////////////////////

	static class FastScanner {
		BufferedReader br;
		StringTokenizer st;

		FastScanner() throws FileNotFoundException {
			br = new BufferedReader(new InputStreamReader(System.in));
		}

		FastScanner(int a) throws FileNotFoundException {
			br = new BufferedReader(new FileReader("input.txt"));
		}

		String next() throws IOException {
			while (st == null || !st.hasMoreElements()) {
				try {
					st = new StringTokenizer(br.readLine());
				} catch (IOException e) {
					e.printStackTrace();
				}
			}
			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());
		}

		String nextLine() throws IOException {
			String str = "";
			try {
				str = br.readLine();
			} catch (IOException e) {
				e.printStackTrace();
			}
			return str;
		}

		int[] readArray(int n) throws IOException {
			int[] A = new int[n];
			for (int i = 0; i != n; i++) {
				A[i] = sc.nextInt();
			}
			return A;
		}

		long[] readArrayLong(int n) throws IOException {
			long[] A = new long[n];
			for (int i = 0; i != n; i++) {
				A[i] = sc.nextLong();
			}
			return A;
		}
	}

	static void ruffleSort(int[] a) {
		Random get = new Random();
		for (int i = 0; i < a.length; i++) {
			int r = get.nextInt(a.length);
			int temp = a[i];
			a[i] = a[r];
			a[r] = temp;
		}
		Arrays.sort(a);
	}

	static void ruffleSort(long[] a) {
		Random get = new Random();
		for (int i = 0; i < a.length; i++) {
			int r = get.nextInt(a.length);
			long temp = a[i];
			a[i] = a[r];
			a[r] = temp;
		}
		Arrays.sort(a);
	}
}