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:
//package Codeforces;

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

public class F2 {
    static int getNextPowerOf2(int n)
    {
        n |= (n>>1);
        n |= (n>>2);
        n |= (n>>4);
        n |= (n>>8);
        n |= (n>>16);
        n |= (n>>30);
        return n+1;
    }
    public static void main(String[] args) throws IOException {
        Soumit sc = new Soumit();

        int n = sc.nextInt();
        int t = sc.nextInt();

        int sn = 2*getNextPowerOf2(n)-1;
        int[] segTree = new int[sn];
        for(int i=0;i<sn;i++){
            segTree[i] = -1;
        }

        while(t-->0){
            int k = sc.nextInt();
            int ll = 0;
            int ul = n-1;
            int prev = 0;
            Stack<Integer> stk = new Stack<>();
            int si = 1;
            while (ll<ul){
                int mid = (ll+ul)/2;

                int sum;
                //System.out.println("Test: "+ll+" "+ul+" "+si);
                if(segTree[si]==-1){
                    System.out.println("? "+(ll+1)+" "+(mid+1));
                    System.out.flush();
                    sum = sc.nextInt();
                    segTree[si] = sum;
                }
                else{
                    sum = segTree[si];
                }

                int range = mid-ll+1;
                int zeros = range-sum;

                //System.out.println(prev+" "+zeros);
                if(prev+zeros>=k){
                    ul = mid;
                    stk.push(si);
                    si = 2*si + 1;
                }
                else{
                    ll = mid+1;
                    prev += zeros;
                    si = si+1;
                    stk.push(si);
                    si = 2*si+1;
                }
            }

            System.out.println("! "+(ll+1));
            while(!stk.isEmpty()){
                int sval = stk.pop();
                segTree[sval] += 1;
            }
        }

        sc.close();
    }

    static class Soumit {
        final private int BUFFER_SIZE = 1 << 18;
        final private DataInputStream din;
        final private byte[] buffer;
        private PrintWriter pw;
        private int bufferPointer, bytesRead;
        StringTokenizer st;

        public Soumit() {
            din = new DataInputStream(System.in);
            buffer = new byte[BUFFER_SIZE];
            bufferPointer = bytesRead = 0;
        }

        public Soumit(String file_name) throws IOException {
            din = new DataInputStream(new FileInputStream(file_name));
            buffer = new byte[BUFFER_SIZE];
            bufferPointer = bytesRead = 0;
        }

        public void streamOutput(String file) throws IOException {
            FileWriter fw = new FileWriter(file);
            BufferedWriter bw = new BufferedWriter(fw);
            pw = new PrintWriter(bw);
        }

        public void println(String a) {
            pw.println(a);
        }

        public void print(String a) {
            pw.print(a);
        }

        public String readLine() throws IOException {
            byte[] buf = new byte[3000064]; // line length
            int cnt = 0, c;
            while ((c = read()) != -1) {
                if (c == '\n')
                    break;
                buf[cnt++] = (byte) c;
            }
            return new String(buf, 0, cnt);
        }

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

        public void sort(int[] arr) {
            ArrayList<Integer> arlist = new ArrayList<>();
            for (int i : arr)
                arlist.add(i);

            Collections.sort(arlist);
            for (int i = 0; i < arr.length; i++)
                arr[i] = arlist.get(i);
        }

        public void sort(long[] arr) {
            ArrayList<Long> arlist = new ArrayList<>();
            for (long i : arr)
                arlist.add(i);

            Collections.sort(arlist);
            for (int i = 0; i < arr.length; i++)
                arr[i] = arlist.get(i);
        }

        public int[] nextIntArray(int n) throws IOException {
            int[] arr = new int[n];
            for (int i = 0; i < n; i++) {
                arr[i] = nextInt();
            }

            return arr;
        }

        public long[] nextLongArray(int n) throws IOException {
            long[] arr = new long[n];
            for (int i = 0; i < n; i++) {
                arr[i] = nextLong();
            }

            return arr;
        }

        public double[] nextDoubleArray(int n) throws IOException {
            double[] arr = new double[n];
            for (int i = 0; i < n; i++) {
                arr[i] = nextDouble();
            }

            return arr;
        }

        public int nextInt() throws IOException {
            int ret = 0;
            byte c = read();
            while (c <= ' ')
                c = read();
            boolean neg = (c == '-');
            if (neg)
                c = read();
            do {
                ret = ret * 10 + c - '0';
            } while ((c = read()) >= '0' && c <= '9');

            if (neg)
                return -ret;
            return ret;
        }

        public long nextLong() throws IOException {
            long ret = 0;
            byte c = read();
            while (c <= ' ')
                c = read();
            boolean neg = (c == '-');
            if (neg)
                c = read();
            do {
                ret = ret * 10 + c - '0';
            }
            while ((c = read()) >= '0' && c <= '9');
            if (neg)
                return -ret;
            return ret;
        }

        public double nextDouble() throws IOException {
            double ret = 0, div = 1;
            byte c = read();
            while (c <= ' ')
                c = read();
            boolean neg = (c == '-');
            if (neg)
                c = read();

            do {
                ret = ret * 10 + c - '0';
            }
            while ((c = read()) >= '0' && c <= '9');

            if (c == '.') {
                while ((c = read()) >= '0' && c <= '9') {
                    ret += (c - '0') / (div *= 10);
                }
            }

            if (neg)
                return -ret;
            return ret;
        }

        private void fillBuffer() throws IOException {
            bytesRead = din.read(buffer, bufferPointer = 0, BUFFER_SIZE);
            if (bytesRead == -1)
                buffer[0] = -1;
        }

        private byte read() throws IOException {
            if (bufferPointer == bytesRead)
                fillBuffer();
            return buffer[bufferPointer++];
        }

        public void close() throws IOException {
            /*if (din == null)
                return;*/
            if (din != null) din.close();
            if (pw != null) pw.close();
        }
    }
}