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:
//package lovejava;
import java.io.*;

import java.math.BigInteger;
import java.util.*;


public class LoveJava {
   static int sum1=0;
    static int sum2=0;
    public static void main(String[] args) {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        PrintWriter pw = new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out)));
        //String s;
        String arr[];
        String arr2[];


        try {
            int t=Integer.parseInt(br.readLine());
            while (t-->0) {
               // int m=Integer.parseInt(br.readLine());

                arr = (br.readLine().replaceAll("//s+$", " ").split(" "));
                arr2 = (br.readLine().replaceAll("//s+$", " ").split(" "));
               // pw.println(game(arr,arr2,m));
                pw.println(func(arr2));

            }
                    pw.close();

                } catch (IOException ex) {
            System.out.println("input mismatch");
        }
    }
    public static long game(String[] arr1,String[] arr2,int m){
        if(m==1)
            return 0;
        int[][]dp=new int[2][m];
        Arrays.fill(dp[0],0);
        Arrays.fill(dp[1],0);
        for(int i=1;i<m;i++){//row1
            dp[0][i]=Integer.parseInt(arr2[i-1])+dp[0][i-1];
        }
        for(int i=1;i<m;i++){//row2
            dp[1][m-1-i]=Integer.parseInt(arr1[m-i])+dp[1][m-i];
        }
        long min=100000;
        for(int i=0;i<m;i++){
            min=min<(Math.max(dp[0][i],dp[1][i]))?min:(Math.max(dp[0][i],dp[1][i]));
        }
        return min;
    }
    public static long func(String[] arr2) {
        int i = 0;
        int[] maxima = new int[arr2.length];
        int[] minima = new int[arr2.length];
        int maxCount = 0, minCount = 0;

        while (i < arr2.length - 1) {
            if(i+1<arr2.length&&Integer.parseInt(arr2[i])==Integer.parseInt(arr2[i+1])){
                i++;
            }
            while (i+1<arr2.length&&Integer.parseInt(arr2[i]) < Integer.parseInt(arr2[i + 1])) {//maxima
                i++;
            }
            maxima[maxCount++] = Integer.parseInt(arr2[i]);
            i++;
            if(i>=arr2.length)
                break;
            while (i+1<arr2.length&&Integer.parseInt(arr2[i]) > Integer.parseInt(arr2[i + 1])){
                i++;
            }
            minima[minCount++]=Integer.parseInt(arr2[i]);
            i++;

        }
        if(arr2.length>=2&&Integer.parseInt(arr2[arr2.length-1])>Integer.parseInt(arr2[arr2.length-2]))
            maxima[maxCount++] = Integer.parseInt(arr2[arr2.length-1]);
        if(maxCount==minCount&&minCount==0)//constant graph
            return Integer.parseInt(arr2[0]);
        else {
            i=0;
            long res=0l;
            res+= (long) maxima[0];
            while (i<maxCount&&i<minCount){
                if(i+1<maxCount&&res-minima[i]+maxima[i+1]>res){
                    res=res-minima[i]+maxima[i+1];

                }
                i++;
            }
            return (long)res;
        }

    }
    /*
    public static int frogJump(int n,int d,String s){
        int count=0;
        int i=0,j;
        for( j=d;j>=1&&i<n;j--){
            if(i+j<n&&(s.charAt(i+j)=='1')) {
                i = i + j;
                count++;
                j=d+1;
            }
        }
        if(i==n-1)
            return count;
        else return -1;
    }
    /*
    public static int Win(String s){
        int i;
        for(i=0;i<=s.length()/2;i++){
            if(s.charAt(i)!=s.charAt(s.length()-1-i))
                break;
        }
        if(i>s.length()/2)
            return 1;//first
        else {
            if((s.length()&1)==0)//even
                return 0;
            else return 1;
        }
    }

     */
    /*
    public static int Win(String[]arr,int n){
        int cnt1=0;
        for(int i=0;i<n;i++) {
            if (arr[i].equals("1"))
                cnt1++;
            if(Integer.parseInt(arr[i])>1)
                break;
        }

            if((cnt1&1)==1) //odd
                if(cnt1==n)
                return 1;//first
                else return 0;//sec
        else {
            if(cnt1==n)
                return 0;
            else return 1;
            }

    }
    */
    /*
    public static int Win(String[]arr,int n,int i){
        if(n==1)
            return 1;//current player won
        else {
            if(Integer.parseInt(arr[i])>1){
                if(Win(arr,n-1,i+1)==1)//leaving one stone
                    return 1;
                else if(Win(arr,n-1,i+1)==0)
                return 1;
                else return 0;
            }
            if(Win(arr,n-1,i+1)==1)
                return 0;
            else return 1;
        }
    }
    */
    /*
    public static int Win(String s){
        int numCnt1=0,numCnt0=0,min=0;
        for(int i=0;i<s.length();i++){
            if(s.charAt(i)=='1')
                numCnt1++;
            else numCnt0++;
        }
        min=Math.min(numCnt0,numCnt1);
        if ((min&1)==1)//odd
            return 1;
        else return 0;
    }

     */
    /*
    public static int Win(String s){//try the other way other sol involving counting number of zeroes and number of 1s
        if(s.equals("01")||s.equals("10"))
            return 1;//win for current player
        if(s.length()==2)
            return 0;
        if(s.length()==1)
            return 0;
        StringBuilder stb=new StringBuilder(s);
            for(int i=0;i<s.length()-1;i++){
                if((s.charAt(i)=='0'&&s.charAt(i+1)=='1')||(s.charAt(i)=='1'&&s.charAt(i+1)=='0')) {
                    String snew=stb.toString();
                    stb.delete(i,i+2);
                   if( Win(stb.toString())==0)
                       return 1;
                   s=snew;
                   stb=new StringBuilder(s);
                }
            }
        return 0;
    }
    */
   /*
    static int find(int N){
        // code here
        String st="";
        StringBuilder s=new StringBuilder(String.valueOf(N));

        for(int i=1;i<s.length();i++){
            if(Integer.parseInt(String.valueOf(s.charAt(i)))>=Integer.parseInt(String.valueOf(s.charAt(i-1))))
                st+=s.charAt(i-1);

            else if(Integer.parseInt(String.valueOf(s.charAt(i)))<Integer.parseInt(String.valueOf(s.charAt(i-1)))) {
                st += String.valueOf(Integer.parseInt(String.valueOf(s.charAt(i - 1))) - 1);
                while (i<s.length()) {
                    st += "9";
                    i++;
                }
            }
            if(i==s.length()-1)
                st+=s.charAt(i);
        }
        return Integer.parseInt(st);
    }
    /*
    public static List<Integer> maxSubarray(List<Integer> arr) {
        // Write your code here
    int res=-10000;
    int sum=0;
    for(Integer e:arr){
        sum+=e;
        if (e>sum)
            sum=e;
        if(sum>res){
            res=sum;
        }
    }
    Collections.sort(arr);
    Collections.reverse(arr);
    int sum2=0;
    int res2=-10000;
    for(Integer e:arr){
        sum2+=e;
        if(sum2>res2)
            res2=sum2;
    }
    ArrayList<Integer> List= new ArrayList<Integer>();
    List.add(res);
    List.add(res2);
    return List;
    }
    }
    */
   }
class Solution
{
    public long nthStair(int n)
    {
        // Code here
        long []dp=new long[n+1];
        Arrays.fill(dp,0l);
        dp[0]=1l;
        for(int i=1;i<=n;i++){
            dp[i]+=dp[i-1];
        }
        for(int i=2;i<=n;i++){
            dp[i]+=dp[i-2];
        }
        return dp[n];
    }
}