Problem: This is an interactive problem!

An arithmetic progression or arithmetic sequence is a sequence of integers such that the subtraction of element with its previous element (x_i - x_{i - 1}, where i ≥ 2) is constant — such difference is called a common difference of the sequence.

That is, an arithmetic progression is a sequence of form x_i = x_1 + (i - 1) d, where d is a common difference of the sequence.

There is a secret list of n integers a_1, a_2, …, a_n.

It is guaranteed that all elements a_1, a_2, …, a_n are between 0 and 10^9, inclusive.

This list is special: if sorted in increasing order, it will form an arithmetic progression with positive common difference (d > 0). For example, the list [14, 24, 9, 19] satisfies this requirement, after sorting it makes a list [9, 14, 19, 24], which can be produced as x_n = 9 + 5 ⋅ (n - 1).

Also you are also given a device, which has a quite discharged battery, thus you can only use it to perform at most 60 queries of following two types:

  * Given a value i (1 ≤ i ≤ n), the device will show the value of the a_i.
  * Given a value x (0 ≤ x ≤ 10^9), the device will return 1 if an element with a value strictly greater than x exists, and it will return 0 otherwise.



Your can use this special device for at most 60 queries. Could you please find out the smallest element and the common difference of the sequence? That is, values x_1 and d in the definition of the arithmetic progression. Note that the array a is not sorted.

Interaction

The interaction starts with a single integer n (2 ≤ n ≤ 10^6), the size of the list of integers.

Then you can make queries of two types:

  * "? i" (1 ≤ i ≤ n) — to get the value of a_i.
  * "> x" (0 ≤ x ≤ 10^9) — to check whether there exists an element greater than x



After the query read its result r as an integer.

  * For the first query type, the r satisfies 0 ≤ r ≤ 10^9. 
  * For the second query type, the r is either 0 or 1.
  * In case you make more than 60 queries or violated the number range in the queries, you will get a r = -1.
  * If you terminate after receiving the -1, you will get the "Wrong answer" verdict. Otherwise you can get an arbitrary verdict because your solution will continue to read from a closed stream. 



When you find out what the smallest element x_1 and common difference d, print

  * "! x_1 d" 



And quit after that. This query is not counted towards the 60 queries limit.

After printing any query do not forget to output end of line and flush the output. Otherwise you will get Idleness limit exceeded. To do this, use:

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



Hacks

For hack, use the following format:

The first line should contain an integer n (2 ≤ n ≤ 10^6) — the list's size.

The second line contains n integers a_1, a_2, …, a_n (0 ≤ a_i ≤ 10^9) — the elements of the list.

Also, after the sorting the list must form an arithmetic progression with positive common difference.

Example

Input


4

0

1

14

24

9

19



Output



&gt; 25

&gt; 15

? 1

? 2

? 3

? 4

! 9 5

Note

Note that the example interaction contains extra empty lines so that it's easier to read. The real interaction doesn't contain any empty lines and you shouldn't print any extra empty lines as well.

The list in the example test is [14, 24, 9, 19].

Solution (in JAVA):

Solution:
// package CF1114;

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


public class Cf1114E {
    static FastReader s;
    static PrintWriter out;
    static String INPUT = "4\n" +
            "\n" +
            "0\n" +
            "\n" +
            "1\n" +
            "\n" +
            "14\n" +
            "\n" +
            "24\n" +
            "\n" +
            "9\n" +
            "\n" +
            "19\n" +
            "\n";

    public static void main(String[] args) {
        long time = System.currentTimeMillis();
        boolean oj = System.getProperty("ONLINE_JUDGE") != null;
        out = new PrintWriter(System.out);
        s = new FastReader(oj);

        int n = s.nextInt();
//        int[] arr = s.nextIntArray(n);
        int l = 0, r = 1000_000_000, max = 0, q = 60;
        while(l <= r) {
            int x = (l + r) >> 1;
            if(q1(x) == 0) {
                max = x;
                r = x - 1;
            } else {
                l = x + 1;
            }
            q--;
        }
        int[] arr = new int[Math.min(q, n)];
        Random rnd = new Random();
        HashSet<Integer> set = new HashSet<>();
        for(int i = 0; i < arr.length; i++) {
            int ind = rnd.nextInt(n) + 1;
            while(set.contains(ind)) {
                ind = rnd.nextInt(n) + 1;
            }
            set.add(ind);
            arr[i] = q2(ind);
        }
        arr = s.uniq(arr);
        int d = 0;
        for(int i = 1; i < arr.length; i++) {
            int diff = arr[i] - arr[i - 1];
            d = Maths.gcd(d, diff);
        }

        int min = max - (n - 1) * d;

        out.println("! " + min + " " + d);
        out.flush();

        if (!oj) {
            System.out.println(Arrays.deepToString(new Object[]{System.currentTimeMillis() - time + " ms"}));
        }
        out.flush();
    }

    private static int q1(int x) {
        out.println("> " + x);
        out.flush();

        return s.nextInt();
    }

    private static int q2(int x) {
        out.println("? " + x);
        out.flush();
        return s.nextInt();
    }


    private static class Maths {
        static ArrayList<Long> printDivisors(long n) {
            // Note that this loop runs till square root
            ArrayList<Long> list = new ArrayList<>();
            for (long i = 1; i <= Math.sqrt(n); i++) {
                if (n % i == 0) {
                    if (n / i == i) {
                        list.add(i);
                    } else {
                        list.add(i);
                        list.add(n / i);
                    }
                }
            }
            return list;
        }


        //    GCD - Using Euclid theorem.
        private static int gcd(int a, int b) {
            if (b == 0) {
                return a;
            }

            return gcd(b, a % b);
        }

        //    Extended euclidean algorithm
//    Used to solve equations of the form ax + by = gcd(a,b)
//    return array [d, a, b] such that d = gcd(p, q), ap + bq = d
        static long[] extendedEuclidean(long p, long q) {
            if (q == 0)
                return new long[]{p, 1, 0};

            long[] vals = extendedEuclidean(q, p % q);
            long d = vals[0];
            long a = vals[2];
            long b = vals[1] - (p / q) * vals[2];
            return new long[]{d, a, b};
        }

        // X ^ y mod p
        static long power(long x, long y, long p) {
            long res = 1;
            x = x % p;

            while (y > 0) {
                if ((y & 1) == 1)
                    res = (res * x) % p;
                y = y >> 1;
                x = (x * x) % p;
            }
            return res;
        }

        // Returns modulo inverse of a
        // with respect to m using extended
        // Euclid Algorithm. Refer below post for details:
        // https://www.geeksforgeeks.org/multiplicative-inverse-under-modulo-m/
        static long inv(long a, long m) {
            long m0 = m, t, q;
            long x0 = 0, x1 = 1;

            if (m == 1)
                return 0;

            // Apply extended Euclid Algorithm
            while (a > 1) {
                q = a / m;
                t = m;
                m = a % m;
                a = t;
                t = x0;
                x0 = x1 - q * x0;
                x1 = t;
            }

            // Make x1 positive
            if (x1 < 0)
                x1 += m0;

            return x1;
        }

        // k is size of num[] and rem[].
        // Returns the smallest number
        // x such that:
        // x % num[0] = rem[0],
        // x % num[1] = rem[1],
        // ..................
        // x % num[k-2] = rem[k-1]
        // Assumption: Numbers in num[] are pairwise
        // coprime (gcd for every pair is 1)
        static long findMinX(long num[], long rem[], long k) {
            int prod = 1;
            for (int i = 0; i < k; i++)
                prod *= num[i];

            int result = 0;
            for (int i = 0; i < k; i++) {
                long pp = prod / num[i];
                result += rem[i] * inv(pp, num[i]) * pp;
            }

            return result % prod;
        }
    }

    private static class BS {
        //  Binary search
        private static int binarySearch(int[] arr, int ele) {
            int low = 0;
            int high = arr.length - 1;
            while (low <= high) {
                int mid = (low + high) / 2;
                if (arr[mid] == ele) {
                    return mid;
                } else if (ele < arr[mid]) {
                    high = mid - 1;
                } else {
                    low = mid + 1;
                }
            }
            return -1;
        }

        //    First occurence using binary search
        private static int binarySearchFirstOccurence(int[] arr, int ele) {
            int low = 0;
            int high = arr.length - 1;
            int ans = -1;
            while (low <= high) {
                int mid = (low + high) / 2;
                if (arr[mid] == ele) {
                    ans = mid;
                    high = mid - 1;
                } else if (ele < arr[mid]) {
                    high = mid - 1;
                } else {
                    low = mid + 1;
                }
            }
            return ans;
        }


        //    Last occurenece using binary search
        private static int binarySearchLastOccurence(int[] arr, int ele) {
            int low = 0;
            int high = arr.length - 1;
            int ans = -1;
            while (low <= high) {
                int mid = (low + high) / 2;
                if (arr[mid] == ele) {
                    ans = mid;
                    low = mid + 1;
                } else if (ele < arr[mid]) {
                    high = mid - 1;
                } else {
                    low = mid + 1;
                }
            }
            return ans;
        }
    }

    private static class arrays {
        // Merge sort
        static void merge(int arr[], int l, int m, int r) {
            int n1 = m - l + 1;
            int n2 = r - m;
            int L[] = new int[n1];
            int R[] = new int[n2];
            for (int i = 0; i < n1; ++i)
                L[i] = arr[l + i];
            for (int j = 0; j < n2; ++j)
                R[j] = arr[m + 1 + j];
            int i = 0, j = 0;
            int k = l;
            while (i < n1 && j < n2) {
                if (L[i] <= R[j]) {
                    arr[k] = L[i];
                    i++;
                } else {
                    arr[k] = R[j];
                    j++;
                }
                k++;
            }

            while (i < n1) {
                arr[k] = L[i];
                i++;
                k++;
            }

            while (j < n2) {
                arr[k] = R[j];
                j++;
                k++;
            }
        }

        static void sort(int arr[], int l, int r) {
            if (l < r) {
                int m = (l + r) / 2;
                sort(arr, l, m);
                sort(arr, m + 1, r);
                merge(arr, l, m, r);
            }
        }

        static void sort(int[] arr) {
            sort(arr, 0, arr.length - 1);
        }
    }

    private static class UnionFindDisjointSet {
        int[] parent;
        int[] size;
        int n;
        int size1;

        public UnionFindDisjointSet(int n) {
            this.n = n;
            this.parent = new int[n];
            this.size = new int[n];
            for (int i = 0; i < n; i++) {
                parent[i] = i;
            }

            for (int i = 0; i < n; i++) {
                size[i] = 1;
            }

            this.size1 = n;
        }

        private int numDisjointSets() {
            System.out.println(size1);
            return size1;
        }

        private boolean find(int a, int b) {
            int rootA = root(a);
            int rootB = root(b);
            if (rootA == rootB) {
                return true;
            }
            return false;
        }

        private int root(int b) {
            while (parent[b] != b) {
                parent[b] = parent[parent[b]];
                b = parent[b];
            }
            return b;
        }

        private void union(int a, int b) {
            int rootA = root(a);
            int rootB = root(b);
            if (rootA == rootB) {
                return;
            }

            if (size[rootA] < size[rootB]) {
                parent[rootA] = parent[rootB];
                size[rootB] += size[rootA];
            } else {
                parent[rootB] = parent[rootA];
                size[rootA] += size[rootB];
            }
            size1--;
            System.out.println(Arrays.toString(parent));
        }
    }

    private static class SegTree {
        int[] st;
        int[] arr;

        public SegTree(int[] arr) {
            this.arr = arr;
            int size = (int) Math.ceil(Math.log(arr.length) / Math.log(2));
            st = new int[(int) ((2 * Math.pow(2, size)) - 1)];
            buildSegmentTree(1, 0, arr.length - 1);
        }

        //**********JUST CALL THE CONSTRUCTOR, THIS FUNCTION WILL BE CALLED AUTOMATICALLY******
        private void buildSegmentTree(int index, int L, int R) {
            if (L == R) {
                st[index] = arr[L];
                return;
            }

            buildSegmentTree(index * 2, L, (L + R) / 2);
            buildSegmentTree(index * 2 + 1, (L + R) / 2 + 1, R);
//          Replace this line if you want to change the function of the Segment tree.
            st[index] = Math.min(st[index * 2], st[index * 2 + 1]);
        }

        //***********We have to use this function **************
        private int Query(int queL, int queR) {
            return Query1(1, 0, arr.length - 1, queL, queR);
        }

        //This is a helper function.
        //************* DO NOT USE THIS ****************
        private int Query1(int index, int segL, int segR, int queL, int queR) {
            if (queL > segR || queR < segL) {
                return -1;
            }

            if (queL <= segL && queR >= segR) {
                return st[index];
            }

            int ans1 = Query1(index * 2, segL, (segL + segR) / 2, queL, queR);
            int ans2 = Query1(index * 2 + 1, (segL + segR) / 2 + 1, segR, queL, queR);
            if (ans1 == -1) {
                return ans2;
            }

            if (ans2 == -1) {
                return ans1;
            }
//             Segment tree implemented for range minimum query. Change the below line to change the function.
            return Math.min(ans1, ans2);
        }

        private void update(int idx, int val) {
            update1(1, 0, arr.length - 1, idx, val);
        }

        private void update1(int node, int queL, int queR, int idx, int val) {
            // idx - index to be updated in the array
            // node - index to be updated in the seg tree
            if (queL == queR) {
                // Leaf node
                arr[idx] += val;
                st[node] += val;
            } else {
                int mid = (queL + queR) / 2;
                if (queL <= idx && idx <= mid) {
                    // If idx is in the left child, recurse on the left child
                    update1(2 * node, queL, mid, idx, val);
                } else {
                    // if idx is in the right child, recurse on the right child
                    update1(2 * node + 1, mid + 1, queR, idx, val);
                }
                // Internal node will have the min of both of its children
                st[node] = Math.min(st[2 * node], st[2 * node + 1]);
            }
        }

    }

    private static class FastReader {
        InputStream is;

        public FastReader(boolean onlineJudge) {
            is = onlineJudge ? System.in : new ByteArrayInputStream(INPUT.getBytes());
        }

        byte[] inbuf = new byte[1024];
        public int lenbuf = 0, ptrbuf = 0;

        int readByte() {
            if (lenbuf == -1)
                throw new InputMismatchException();
            if (ptrbuf >= lenbuf) {
                ptrbuf = 0;
                try {
                    lenbuf = is.read(inbuf);
                } catch (IOException e) {
                    throw new InputMismatchException();
                }
                if (lenbuf <= 0)
                    return -1;
            }
            return inbuf[ptrbuf++];
        }

        boolean isSpaceChar(int c) {
            return !(c >= 33 && c <= 126);
        }

        int skip() {
            int b;
            while ((b = readByte()) != -1 && isSpaceChar(b))
                ;
            return b;
        }

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

        char nextChar() {
            return (char) skip();
        }

        String next() {
            int b = skip();
            StringBuilder sb = new StringBuilder();
            while (!(isSpaceChar(b))) { // when nextLine, (isSpaceChar(b) && b != ' ')
                sb.appendCodePoint(b);
                b = readByte();
            }
            return sb.toString();
        }

        String nextLine() {
            int b = skip();
            StringBuilder sb = new StringBuilder();
            while ((!isSpaceChar(b) || b == ' ')) { // when nextLine, (isSpaceChar(b) && b != ' ')
                sb.appendCodePoint(b);
                b = readByte();
            }
            return sb.toString();
        }

        char[] next(int n) {
            char[] buf = new char[n];
            int b = skip(), p = 0;
            while (p < n && !(isSpaceChar(b))) {
                buf[p++] = (char) b;
                b = readByte();
            }
            return n == p ? buf : Arrays.copyOf(buf, p);
        }

        int nextInt() {
            int num = 0, b;
            boolean minus = false;
            while ((b = readByte()) != -1 && !((b >= '0' && b <= '9') || b == '-'))
                ;
            if (b == '-') {
                minus = true;
                b = readByte();
            }

            while (true) {
                if (b >= '0' && b <= '9') {
                    num = num * 10 + (b - '0');
                } else {
                    return minus ? -num : num;
                }
                b = readByte();
            }
        }

        long nextLong() {
            long num = 0;
            int b;
            boolean minus = false;
            while ((b = readByte()) != -1 && !((b >= '0' && b <= '9') || b == '-'))
                ;
            if (b == '-') {
                minus = true;
                b = readByte();
            }

            while (true) {
                if (b >= '0' && b <= '9') {
                    num = num * 10 + (b - '0');
                } else {
                    return minus ? -num : num;
                }
                b = readByte();
            }
        }

        char[][] nextMatrix(int n, int m) {
            char[][] map = new char[n][];
            for (int i = 0; i < n; i++)
                map[i] = next(m);
            return map;
        }

        int[] nextIntArray(int n) {
            int[] a = new int[n];
            for (int i = 0; i < n; i++)
                a[i] = nextInt();
            return a;
        }

        long[] nextLongArray(int n) {
            long[] a = new long[n];
            for (int i = 0; i < n; i++)
                a[i] = nextLong();
            return a;
        }

        int[][] next2DInt(int n, int m) {
            int[][] arr = new int[n][];
            for (int i = 0; i < n; i++) {
                arr[i] = nextIntArray(m);
            }
            return arr;
        }

        long[][] next2DLong(int n, int m) {
            long[][] arr = new long[n][];
            for (int i = 0; i < n; i++) {
                arr[i] = nextLongArray(m);
            }
            return arr;
        }

        int[] shuffle(int[] arr) {
            Random r = new Random();
            for (int i = 1, j; i < arr.length; i++) {
                j = r.nextInt(i);
                arr[i] = arr[i] ^ arr[j];
                arr[j] = arr[i] ^ arr[j];
                arr[i] = arr[i] ^ arr[j];
            }
            return arr;
        }

        int[] uniq(int[] arr) {
            Arrays.sort(arr);
            int[] rv = new int[arr.length];
            int pos = 0;
            rv[pos++] = arr[0];
            for (int i = 1; i < arr.length; i++) {
                if (arr[i] != arr[i - 1]) {
                    rv[pos++] = arr[i];
                }
            }
            return Arrays.copyOf(rv, pos);
        }
    }

}