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.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.OutputStream;
import java.io.IOException;
import java.io.UncheckedIOException;
import java.io.Closeable;
import java.io.Writer;
import java.io.OutputStreamWriter;
import java.io.InputStream;

/**
 * @author rozerhacks
 */
public class Main {
    public static void main(String[] args) throws Exception {
        Thread thread = new Thread(null, new TaskAdapter(), "rozerhacks", 1 << 29);
        thread.start();
        thread.join();
    }

    static class TaskAdapter implements Runnable {
        @Override
        public void run() {
            InputStream inputStream = System.in;
            OutputStream outputStream = System.out;
            FastInput in = new FastInput(inputStream);
            FastOutput out = new FastOutput(outputStream);
            C2PokmonArmyHardVersion solver = new C2PokmonArmyHardVersion();
            int testCount = Integer.parseInt(in.next());
            for (int i = 1; i <= testCount; i++)
                solver.solve(i, in, out);
            out.close();
        }
    }

    static class C2PokmonArmyHardVersion {
        long[][] tree;
        int[] ar;

        void build(int node, int start, int end) {
            if (start == end) {
                tree[node][0] = ar[start];
                tree[node][1] = Long.MIN_VALUE / 10;
                tree[node][2] = Long.MIN_VALUE / 10;
                tree[node][3] = -ar[start];
            } else {
                int mid = (start + end) >> 1;
                build(node << 1, start, mid);
                build(node << 1 | 1, mid + 1, end);
                for (int i = 0; i < 4; i++) tree[node][i] = Math.max(tree[node << 1][i], tree[node << 1 | 1][i]);

                tree[node][0] = Math.max(tree[node][0], tree[node << 1][0] + tree[node << 1 | 1][2]);
                tree[node][0] = Math.max(tree[node][0], tree[node << 1][1] + tree[node << 1 | 1][0]);

                tree[node][1] = Math.max(tree[node][1], tree[node << 1][1] + tree[node << 1 | 1][1]);
                tree[node][1] = Math.max(tree[node][1], tree[node << 1][0] + tree[node << 1 | 1][3]);

                tree[node][2] = Math.max(tree[node][2], tree[node << 1][2] + tree[node << 1 | 1][2]);
                tree[node][2] = Math.max(tree[node][2], tree[node << 1][3] + tree[node << 1 | 1][0]);

                tree[node][3] = Math.max(tree[node][3], tree[node << 1][3] + tree[node << 1 | 1][1]);
                tree[node][3] = Math.max(tree[node][3], tree[node << 1][2] + tree[node << 1 | 1][3]);
            }
        }

        void update(int node, int start, int end, int index, int value) {
            if (index > end || index < start) return;
            if (index == start && index == end) {
                tree[node][0] = value;
                tree[node][1] = Long.MIN_VALUE / 10;
                tree[node][2] = Long.MIN_VALUE / 10;
                tree[node][3] = -value;
                return;
            }
            int mid = (start + end) >> 1;
            if (index <= mid) update(node << 1, start, mid, index, value);
            else update(node << 1 | 1, mid + 1, end, index, value);

            for (int i = 0; i < 4; i++) tree[node][i] = Math.max(tree[node << 1][i], tree[node << 1 | 1][i]);
            tree[node][0] = Math.max(tree[node][0], tree[node << 1][0] + tree[node << 1 | 1][2]);
            tree[node][0] = Math.max(tree[node][0], tree[node << 1][1] + tree[node << 1 | 1][0]);

            tree[node][1] = Math.max(tree[node][1], tree[node << 1][1] + tree[node << 1 | 1][1]);
            tree[node][1] = Math.max(tree[node][1], tree[node << 1][0] + tree[node << 1 | 1][3]);

            tree[node][2] = Math.max(tree[node][2], tree[node << 1][2] + tree[node << 1 | 1][2]);
            tree[node][2] = Math.max(tree[node][2], tree[node << 1][3] + tree[node << 1 | 1][0]);

            tree[node][3] = Math.max(tree[node][3], tree[node << 1][3] + tree[node << 1 | 1][1]);
            tree[node][3] = Math.max(tree[node][3], tree[node << 1][2] + tree[node << 1 | 1][3]);
        }

        public void solve(int testNumber, FastInput in, FastOutput out) {
            int n = in.readInt();
            int q = in.readInt();
            ar = new int[n];
            in.populate(ar);
            tree = new long[4 * (n + 5)][4];
            build(1, 0, n - 1);
            out.println(Math.max(tree[1][0], tree[1][1]));
            for (int i = 0; i < q; i++) {
                int l = in.readInt() - 1;
                int r = in.readInt() - 1;
                update(1, 0, n - 1, l, ar[r]);
                update(1, 0, n - 1, r, ar[l]);
                ar[l] = ar[l] + ar[r] - (ar[r] = ar[l]);
                out.println(Math.max(tree[1][0], tree[1][1]));
            }
        }

    }

    static class FastInput {
        private final InputStream is;
        private final StringBuilder defaultStringBuf = new StringBuilder(1 << 13);
        private final byte[] buf = new byte[1 << 13];
        private int bufLen;
        private int bufOffset;
        private int next;

        public FastInput(InputStream is) {
            this.is = is;
        }

        public void populate(int[] data) {
            for (int i = 0; i < data.length; i++) {
                data[i] = readInt();
            }
        }

        private int read() {
            while (bufLen == bufOffset) {
                bufOffset = 0;
                try {
                    bufLen = is.read(buf);
                } catch (IOException e) {
                    bufLen = -1;
                }
                if (bufLen == -1) {
                    return -1;
                }
            }
            return buf[bufOffset++];
        }

        public void skipBlank() {
            while (next >= 0 && next <= 32) {
                next = read();
            }
        }

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

        public int readInt() {
            int sign = 1;

            skipBlank();
            if (next == '+' || next == '-') {
                sign = next == '+' ? 1 : -1;
                next = read();
            }

            int val = 0;
            if (sign == 1) {
                while (next >= '0' && next <= '9') {
                    val = val * 10 + next - '0';
                    next = read();
                }
            } else {
                while (next >= '0' && next <= '9') {
                    val = val * 10 - next + '0';
                    next = read();
                }
            }

            return val;
        }

        public String readString(StringBuilder builder) {
            skipBlank();

            while (next > 32) {
                builder.append((char) next);
                next = read();
            }

            return builder.toString();
        }

        public String readString() {
            defaultStringBuf.setLength(0);
            return readString(defaultStringBuf);
        }

    }

    static class FastOutput implements AutoCloseable, Closeable, Appendable {
        private final Writer os;
        private final StringBuilder cache = new StringBuilder(5 << 20);

        public FastOutput(Writer os) {
            this.os = os;
        }

        public FastOutput(OutputStream os) {
            this(new OutputStreamWriter(os));
        }

        public FastOutput append(CharSequence csq) {
            cache.append(csq);
            return this;
        }

        public FastOutput append(CharSequence csq, int start, int end) {
            cache.append(csq, start, end);
            return this;
        }

        public FastOutput append(char c) {
            cache.append(c);
            return this;
        }

        public FastOutput append(long c) {
            cache.append(c);
            return this;
        }

        public FastOutput println(long c) {
            return append(c).println();
        }

        public FastOutput println() {
            cache.append(System.lineSeparator());
            return this;
        }

        public FastOutput flush() {
            try {
                os.append(cache);
                os.flush();
                cache.setLength(0);
            } catch (IOException e) {
                throw new UncheckedIOException(e);
            }
            return this;
        }

        public void close() {
            flush();
            try {
                os.close();
            } catch (IOException e) {
                throw new UncheckedIOException(e);
            }
        }

        public String toString() {
            return cache.toString();
        }

    }
}