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.lang.*;
import java.io.*;

public class Main
{
    
	public static void main (String[] args) throws java.lang.Exception
	{
		FastReader sc =new FastReader();
		
	    int t=sc.nextInt();
	    
	   // int t=1;
	    
	    while(t-->0)
	    {
	        int n=sc.nextInt();
	        int k=sc.nextInt();
	        int arr[] = sc.readArray(n);
	        long ans = arr[0];
	        
	        for(int i=1;i<n;i++)
	        {
	            ans+=Math.max(0,arr[i]-arr[i-1]);
	        }
	        System.out.println(ans);
	    }
	}
	
	
	
	static class FastReader {
        BufferedReader br;
        StringTokenizer st;
 
        public FastReader()
        {
            br = new BufferedReader(
                new InputStreamReader(System.in));
        }
 
        String next()
        {
            while (st == null || !st.hasMoreElements()) {
                try {
                    st = new StringTokenizer(br.readLine());
                }
                catch (IOException e) {
                    e.printStackTrace();
                }
            }
            return st.nextToken();
        }
 
        int nextInt() { return Integer.parseInt(next()); }
 
        long nextLong() { return Long.parseLong(next()); }
 
        double nextDouble()
        {
            return Double.parseDouble(next());
        }
        
        float nextFloat()
        {
            return Float.parseFloat(next());
        }
 
        String nextLine()
        {
            String str = "";
            try {
                str = br.readLine();
            }
            catch (IOException e) {
                e.printStackTrace();
            }
            return str;
        }
        
        int[] readArray(int n) {
			int[] a=new int[n];
			for (int i=0; i<n; i++) a[i]=nextInt();
			return a;
		}
		
		long[] readArrayLong(int n) {
			long[] a=new long[n];
			for (int i=0; i<n; i++) a[i]=nextLong();
			return a;
		}
		
    }
    
    public static int[] radixSort2(int[] a)
	{
		int n = a.length;
		int[] c0 = new int[0x101];
		int[] c1 = new int[0x101];
		int[] c2 = new int[0x101];
		int[] c3 = new int[0x101];
		for(int v : a) {
			c0[(v&0xff)+1]++;
			c1[(v>>>8&0xff)+1]++;
			c2[(v>>>16&0xff)+1]++;
			c3[(v>>>24^0x80)+1]++;
		}
		for(int i = 0;i < 0xff;i++) {
			c0[i+1] += c0[i];
			c1[i+1] += c1[i];
			c2[i+1] += c2[i];
			c3[i+1] += c3[i];
		}
		int[] t = new int[n];
		for(int v : a)t[c0[v&0xff]++] = v;
		for(int v : t)a[c1[v>>>8&0xff]++] = v;
		for(int v : a)t[c2[v>>>16&0xff]++] = v;
		for(int v : t)a[c3[v>>>24^0x80]++] = v;
		return a;
	}
    
    static boolean isPalindrome(String str)
    {
 
        // Pointers pointing to the beginning
        // and the end of the string
        int i = 0, j = str.length() - 1;
 
        // While there are characters to compare
        while (i < j) {
 
            // If there is a mismatch
            if (str.charAt(i) != str.charAt(j))
                return false;
 
            // Increment first pointer and
            // decrement the other
            i++;
            j--;
        }
 
        // Given string is a palindrome
        return true;
    }
    
    static boolean palindrome_array(int arr[], int n)
    {
        // Initialise flag to zero.
        int flag = 0;
 
        // Loop till array size n/2.
        for (int i = 0; i <= n / 2 && n != 0; i++) {
 
            // Check if first and last element are different
            // Then set flag to 1.
            if (arr[i] != arr[n - i - 1]) {
                flag = 1;
                break;
            }
        }
 
        // If flag is set then print Not Palindrome
        // else print Palindrome.
        if (flag == 1)
            return false;
        else
            return true;
    }
    
	static boolean allElementsEqual(int[] arr,int n)
	{
	    int z=0;
	        for(int i=0;i<n-1;i++)
		    {
		        if(arr[i]==arr[i+1])
		        {
		            z++;
		        }
		    }
		    if(z==n-1)
		    {
		        return true;
		    }
		    else
		    {
		        return false;
		    }
		    
	}
	
	static boolean allElementsDistinct(int[] arr,int n)
	{
	    int z=0;
	        for(int i=0;i<n-1;i++)
		    {
		        if(arr[i]!=arr[i+1])
		        {
		            z++;
		        }
		    }
		    if(z==n-1)
		    {
		        return true;
		    }
		    else
		    {
		        return false;
		    }
		    
		    
	}
	
	public static void reverse(int[] array)
    {
  
        // Length of the array
        int n = array.length;
  
        // Swaping the first half elements with last half
        // elements
        for (int i = 0; i < n / 2; i++) {
  
            // Storing the first half elements temporarily
            int temp = array[i];
  
            // Assigning the first half to the last half
            array[i] = array[n - i - 1];
  
            // Assigning the last half to the first half
            array[n - i - 1] = temp;
        }
    }
    
    public static void reverse_Long(long[] array)
    {
  
        // Length of the array
        int n = array.length;
  
        // Swaping the first half elements with last half
        // elements
        for (int i = 0; i < n / 2; i++) {
  
            // Storing the first half elements temporarily
            long temp = array[i];
  
            // Assigning the first half to the last half
            array[i] = array[n - i - 1];
  
            // Assigning the last half to the first half
            array[n - i - 1] = temp;
        }
    }
	
	static boolean isSorted(int[] a)
    {
        for (int i = 0; i < a.length - 1; i++)
        {
            if (a[i] > a[i + 1]) {
                return false;
            }
        }
 
        return true;
    }
    
    
    
    static boolean isReverseSorted(int[] a)
    {
        for (int i = 0; i < a.length - 1; i++)
        {
            if (a[i] < a[i + 1]) {
                return false;
            }
        }
 
        return true;
    }
    
    static int[] rearrangeEvenAndOdd(int arr[], int n)
    {
        ArrayList<Integer> list = new ArrayList<>();
		    
		    for(int i=0;i<n;i++)
		    {
		        if(arr[i]%2==0)
		        {
		            list.add(arr[i]);
		        }
		    }
		    
		    for(int i=0;i<n;i++)
		    {
		        if(arr[i]%2!=0)
		        {
		            list.add(arr[i]);
		        }
		    }
		    int len = list.size();
		    int[] array = list.stream().mapToInt(i->i).toArray();
		    return array;
    }
    
    static long[] rearrangeEvenAndOddLong(long arr[], int n)
    {
        ArrayList<Long> list = new ArrayList<>();
		    
		    for(int i=0;i<n;i++)
		    {
		        if(arr[i]%2==0)
		        {
		            list.add(arr[i]);
		        }
		    }
		    
		    for(int i=0;i<n;i++)
		    {
		        if(arr[i]%2!=0)
		        {
		            list.add(arr[i]);
		        }
		    }
		    int len = list.size();
		    long[] array = list.stream().mapToLong(i->i).toArray();
		    return array;
    }
	
	
    
    static boolean isPrime(long n)
    {
 
        // Check if number is less than
        // equal to 1
        if (n <= 1)
            return false;
 
        // Check if number is 2
        else if (n == 2)
            return true;
 
        // Check if n is a multiple of 2
        else if (n % 2 == 0)
            return false;
 
        // If not, then just check the odds
        for (long i = 3; i <= Math.sqrt(n); i += 2)
        {
            if (n % i == 0)
                return false;
        }
        return true;
    }
    
    static int getSum(int n)
    {    
        int sum = 0;
          
        while (n != 0)
        {
            sum = sum + n % 10;
            n = n/10;
        }
      
    return sum;
    }
    
    static int gcd(int a, int b)
    {
      if (b == 0)
        return a;
      return gcd(b, a % b);
    }
     
    static long gcdLong(long a, long b)
    {
        if (b == 0)
        return a;
      return gcdLong(b, a % b);
    }
    
    static void swap(int[] arr, int i, int j)
    {
        int temp = arr[i];
        arr[i] = arr[j];
        arr[j] = temp;
    }
	
	static int countDigit(long n)
    {
        return (int)Math.floor(Math.log10(n) + 1);
    }
	
}