Problem: This is the easy version of the problem. The difference between the versions is that the easy version has no swap operations. You can make hacks only if all versions of the problem are solved.

Pikachu is a cute and friendly pokémon living in the wild pikachu herd.

But it has become known recently that infamous team R wanted to steal all these pokémon! Pokémon trainer Andrew decided to help Pikachu to build a pokémon army to resist.

First, Andrew counted all the pokémon — there were exactly n pikachu. The strength of the i-th pokémon is equal to a_i, and all these numbers are distinct.

As an army, Andrew can choose any non-empty subsequence of pokemons. In other words, Andrew chooses some array b from k indices such that 1 ≤ b_1 < b_2 < ... < b_k ≤ n, and his army will consist of pokémons with forces a_{b_1}, a_{b_2}, ..., a_{b_k}.

The strength of the army is equal to the alternating sum of elements of the subsequence; that is, a_{b_1} - a_{b_2} + a_{b_3} - a_{b_4} + ....

Andrew is experimenting with pokémon order. He performs q operations. In i-th operation Andrew swaps l_i-th and r_i-th pokémon.

Note: q=0 in this version of the task.

Andrew wants to know the maximal stregth of the army he can achieve with the initial pokémon placement. He also needs to know the maximal strength after each operation.

Help Andrew and the pokémon, or team R will realize their tricky plan!

Input

Each test contains multiple test cases.

The first line contains one positive integer t (1 ≤ t ≤ 10^3) denoting the number of test cases. Description of the test cases follows.

The first line of each test case contains two integers n and q (1 ≤ n ≤ 3 ⋅ 10^5, q = 0) denoting the number of pokémon and number of operations respectively.

The second line contains n distinct positive integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ n) denoting the strengths of the pokémon.

i-th of the last q lines contains two positive integers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) denoting the indices of pokémon that were swapped in the i-th operation.

It is guaranteed that the sum of n over all test cases does not exceed 3 ⋅ 10^5, and the sum of q over all test cases does not exceed 3 ⋅ 10^5. 

Output

For each test case, print q+1 integers: the maximal strength of army before the swaps and after each swap.

Example

Input


3
3 0
1 3 2
2 0
1 2
7 0
1 2 5 4 3 6 7


Output


3
2
9

Note

In third test case we can build an army in such way: [1 2 5 4 3 6 7], its strength will be 5−3+7=9.

Solution (in JAVA):

Solution:
import java.io.BufferedWriter;
import java.io.IOException;
import java.io.InputStream;
import java.io.OutputStream;
import java.io.OutputStreamWriter;
import java.io.PrintWriter;
import java.io.Writer;
import java.util.ArrayList;
import java.util.HashSet;
import java.util.InputMismatchException;
import java.util.List;
import java.util.Set;

public class Codeforces {

    int a[];
    long ans = 0;

    public void solve2() {
        InputReader in = new InputReader(System.in);
        OutputWriter out = new OutputWriter(System.out);
        int t = in.readInt();
        while (t-- > 0) {
            int n = in.readInt(), q = in.readInt();
            a = new int[n + 2];
            for (int i = 1; i <= n; i++) {
                a[i] = in.readInt();
            }
            a[0] = -1;
            a[n + 1] = -1;
            ans = 0;
            for (int i = 1; i <= n; i++) {
                add(i, 1, n);
            }
            out.printLine(ans);
            while (q-- > 0) {
                int i = in.readInt(), j = in.readInt();
                //revert the additions/subtractions done on i, j and their neighbours.
                queryAdd(i, j, -1, n);
                //swap
                int temp = a[i];
                a[i] = a[j];
                a[j] = temp;
                //again calculate for i, j and their neighbours.
                queryAdd(i, j, 1, n);
                out.printLine(ans);
            }
        }
        out.flush();
    }

    private void queryAdd(int i, int j, int sign, int n) {
        // Each swap will affect only i, j and their neighbours. 
        // So recalculate just the values of i, j and their neighbours. 
        if ( i + 1 == j ) {
            add(i - 1, sign, n);
            add(i, sign, n);
            add(j, sign, n);
            add(j + 1, sign, n);
        } else {
            add(i - 1, sign, n);
            add(i, sign, n);
            add(i + 1, sign, n);
            add(j - 1, sign, n);
            add(j, sign, n);
            add(j + 1, sign, n);
        }
    }

    private void add(int i, int sign, int n) {
        if ( i <= 0 || i >= n + 1 ) return;
        if ( a[i] > a[i - 1] && a[i] > a[i + 1] ) ans += sign * a[i];
        if ( a[i] < a[i + 1] && a[i] < a[i - 1] ) ans -= sign * a[i];
    }

    public static void main(String[] args) {
        Codeforces solver = new Codeforces();
        solver.solve2();
    }

    static class InputReader {

        private InputStream stream;
        private byte[] buf = new byte[1024];
        private int curChar;
        private int numChars;
        private SpaceCharFilter filter;

        public InputReader(InputStream stream) {
            this.stream = stream;
        }

        public int read() {
            if ( numChars == -1 )
                throw new InputMismatchException();
            if ( curChar >= numChars ) {
                curChar = 0;
                try {
                    numChars = stream.read(buf);
                } catch (IOException e) {
                    throw new InputMismatchException();
                }
                if ( numChars <= 0 )
                    return -1;
            }
            return buf[curChar++];
        }

        public int readInt() {
            int c = read();
            while (isSpaceChar(c))
                c = read();
            int sgn = 1;
            if ( c == '-' ) {
                sgn = -1;
                c = read();
            }
            int res = 0;
            do {
                if ( c < '0' || c > '9' )
                    throw new InputMismatchException();
                res *= 10;
                res += c - '0';
                c = read();
            } while (!isSpaceChar(c));
            return res * sgn;
        }

        public long readLong() {
            int c = read();
            while (isSpaceChar(c))
                c = read();
            int sgn = 1;
            if ( c == '-' ) {
                sgn = -1;
                c = read();
            }
            long res = 0;
            do {
                if ( c < '0' || c > '9' )
                    throw new InputMismatchException();
                res *= 10;
                res += c - '0';
                c = read();
            } while (!isSpaceChar(c));
            return res * sgn;
        }

        public String readString() {
            int c = read();
            while (isSpaceChar(c))
                c = read();
            StringBuilder res = new StringBuilder();
            do {
                res.appendCodePoint(c);
                c = read();
            } while (!isSpaceChar(c));
            return res.toString();
        }

        public boolean isSpaceChar(int c) {
            if ( filter != null )
                return filter.isSpaceChar(c);
            return c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == -1;
        }

        public String next() {
            return readString();
        }

        public interface SpaceCharFilter {
            public boolean isSpaceChar(int ch);
        }
    }

    static class OutputWriter {
        private final PrintWriter writer;

        public OutputWriter(OutputStream outputStream) {
            writer = new PrintWriter(new BufferedWriter(new OutputStreamWriter(outputStream)));
        }

        public OutputWriter(Writer writer) {
            this.writer = new PrintWriter(writer);
        }

        public void print(Object... objects) {
            for (int i = 0; i < objects.length; i++) {
                if ( i != 0 )
                    writer.print(' ');
                writer.print(objects[i]);
            }
        }

        public void printLine(Object... objects) {
            print(objects);
            writer.println();
        }

        public void close() {
            writer.close();
        }

        public void flush() {
            writer.flush();
        }

    }

    static class IOUtils {

        public static int[] readIntegerArray(InputReader in, int size) {
            int[] array = new int[size];
            for (int i = 0; i < size; i++) {
                array[i] = in.readInt();
            }
            return array;
        }

        public static long[] readLongArray(InputReader in, int size) {
            long[] array = new long[size];
            for (int i = 0; i < size; i++) {
                array[i] = in.readLong();
            }
            return array;
        }

        public static List<Integer> readIntegerList(InputReader in, int size) {
            List<Integer> set = new ArrayList<>();
            for (int i = 0; i < size; i++) {
                set.add(in.readInt());
            }
            return set;
        }

        public static Set<Integer> readIntegerSet(InputReader in, int size) {
            Set<Integer> set = new HashSet<Integer>();
            for (int i = 0; i < size; i++) {
                set.add(in.readInt());
            }
            return set;
        }
    }
}