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.util.*;
import java.io.*;
import java.math.*;
 
public class Main {
    static PrintWriter out;
    static Reader in;
    public static void main(String[] args) throws IOException {
        //out = new PrintWriter(new File("out.txt"));
        //PrintWriter out = new PrintWriter(System.out);
        //in = new Reader(new FileInputStream("in.txt"));
        //Reader in = new Reader();
        input_output();
        Main solver = new Main();
        solver.solve();
        out.flush();
        out.close();
 
    }
 
    static long INF = (long)1e16;
    static int maxn = (int)1e3+5;
    static int mod=(int)1e9+7;
    static int n, m, q, k, t;
 
    static long ans;

    void solve() throws IOException{
        t = in.nextInt();

        while (t --> 0) {
            n = in.nextInt();
            q = in.nextInt();

            int[] arr = new int[n+2];
            for (int i = 1; i <= n; i++) arr[i] = in.nextInt();
            n++;

            ans = 0;
            int[] sign = new int[n+2];
            sign[0] = 1;
            for (int i = 1; i <= n; i++) {
                if (sign[i-1] == 1) {
                    if (arr[i] > arr[i-1]) sign[i] = 1;
                    else {
                        sign[i] = -1;
                        ans += arr[i-1];
                    }
                } else {
                    if (arr[i] < arr[i-1]) sign[i] = -1;
                    else {
                        sign[i] = 1;
                        ans -= arr[i-1];
                    }
                }
            }
            out.println(ans);

            int l, r, vl, vr;
            for (int i = 0; i < q; i++) {
                l = in.nextInt(); vl = arr[l];
                r = in.nextInt(); vr = arr[r];

                change(arr, sign, l, n+2);
                //out.println((n+2)+" "+l+" "+ans);
                change(arr, sign, r, n+1);
                //out.println((n+1)+" "+r+" "+ans);
                change(arr, sign, l, vr);
                //out.println(vr+" "+l+" "+ans);
                change(arr, sign, r, vl);
                //out.println(vl+" "+r+" "+ans);
                out.println(ans);
            }
        }
    }
 
    //<>

    static void change(int[] arr, int[] sign, int idx, int val) {
        int cur, r = 0;
        boolean dif = sign[idx-1] != sign[idx];
        for (int i = idx; i <= n; i++) {
            cur = sign[i];
            if (i != n && sign[i] != sign[i+1]) {
                if (sign[i] == 1) {
                    ans -= arr[i];
                    //if (idx == 1 && val == 3) System.out.println("henno "+(-arr[i]));
                } else { 
                    ans += arr[i];
                    //if (idx == 1 && val == 3) System.out.println("henno "+arr[i]);
                }
            }
            if (i == idx) arr[idx] = val;
            if (sign[i-1] == 1) {
                if (arr[i] > arr[i-1]) sign[i] = 1;
                else sign[i] = -1;
            } else {
                if (arr[i] < arr[i-1]) sign[i] = -1;
                else sign[i] = 1;
            }

            r = i;
            if (i != idx && cur == sign[i]) break;
        }
        //if (idx == 1 && val == 3) System.out.println("hello "+r);
        for (int i = idx; i <= r; i++) {
            if (i != n && sign[i] != sign[i+1]) {
                if (sign[i] == 1)  {
                    ans += arr[i];
                    //if (idx == 1 && val == 3) System.out.println("hey "+arr[i]);
                } else {
                    //if (idx == 1 && val == 3) System.out.println("hey "+(-arr[i]));
                    ans -= arr[i];
                }
            }
        }
        boolean difnow = sign[idx-1] != sign[idx];
        //if (idx == 1 && val == 3) System.out.println(difnow == dif);
        if (dif != difnow) {
            if (dif) {
                if (sign[idx-1] == 1) ans -= arr[idx-1];
                else ans += arr[idx-1];
            } else {
                if (sign[idx-1] == 1) ans += arr[idx-1];
                else ans -= arr[idx-1];
            }
        }
    }

    static class Reader {
 
        private InputStream mIs;
        private byte[] buf = new byte[1024];
        private int curChar;
        private int numChars;
    
        public Reader() {
            this(System.in);
        }
    
        public Reader(InputStream is) {
            mIs = is;
        }
    
        public int read() {
            if (numChars == -1) {
                throw new InputMismatchException();
    
        }
            if (curChar >= numChars) {
                curChar = 0;
                try {
                    numChars = mIs.read(buf);
                } catch (IOException e) {
                    throw new InputMismatchException();
                }
                if (numChars <= 0) {
                    return -1;
                }
            }
            return buf[curChar++];
        }
    
        public String nextLine() {
            int c = read();
            while (isSpaceChar(c)) {
                c = read();
            }
            StringBuilder res = new StringBuilder();
            do {
                res.appendCodePoint(c);
                c = read();
            } while (!isEndOfLine(c));
            return res.toString();
        }
    
        public String next() {
            int c = read();
            while (isSpaceChar(c)) {
                c = read();
            }
            StringBuilder res = new StringBuilder();
            do {
                res.appendCodePoint(c);
                c = read();
            } while (!isSpaceChar(c));
            return res.toString();
        }
    
        double nextDouble()
        {
            return Double.parseDouble(next());
        }
    
        public long nextLong() {
            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 int nextInt() {
            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 boolean isSpaceChar(int c) {
            return c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == -1;
        }
    
        public boolean isEndOfLine(int c) {
            return c == '\n' || c == '\r' || c == -1;
        }
 
    }
    static void input_output() throws IOException {
        File f = new File("in.txt");
        if(f.exists() && !f.isDirectory()) { 
            in = new Reader(new FileInputStream("in.txt"));
        } else in = new Reader();
        f = new File("out.txt");
        if(f.exists() && !f.isDirectory()) {
            out = new PrintWriter(new File("out.txt"));
        } else out = new PrintWriter(System.out);
    }
}