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.*;
import java.util.*;

public class Main {
	static ArrayList<Integer> adj[];

	static int k;
	static long[] a;
	static int b[];
	static int m;

	static class Pair implements Comparable<Pair> {
		int c;
		int d;
		public Pair(int b, int l) {
			c = b;
			d = l;
		}
		@Override
		public int compareTo(Pair o) {
		return o.d-this.d;
		}



	}

	static ArrayList<Pair> adjlist[];
	// static char c[];
	static long mod = (long) (998244353);
	static int V;
	static long INF = (long) 1E16;
	static int n;
	static char c[];
	static Pair p[];
static long grid[][];
static Scanner sc = new Scanner(System.in);
static PrintWriter out = new PrintWriter(System.out);
public static void main(String args[]) throws Exception {
	int t=sc.nextInt();
	while(t-->0) {
	int n=sc.nextInt();
	int q=sc.nextInt();
	a=new long[n];
	total=0;
	for (int i = 0; i < a.length; i++) {
		update(i,sc.nextInt() );
	
	}
	out.println(total);
	while(q-->0) {
		int l=sc.nextInt()-1;
		int r=sc.nextInt()-1;
		long temp=a[l];
		update(l, (int) a[r]);
		update(r, (int)temp);
		out.println(total);
	}
	}
	out.flush();
}
static void update(int idx,int val) {
	if(idx==0) {
		total-=a[0];
		if(idx+1<a.length)
		total-=Math.max(0,a[idx+1]-a[idx]);
		a[0]=val;
		total+=a[0];
		if(idx+1<a.length)
		total+=Math.max(0,a[idx+1]-a[idx]);
		
		return;
	}
	if(idx>0) {
		total-=Math.max(0,a[idx]-a[idx-1]);
	}
	 if(idx+1<a.length) {
		total-=Math.max(0,a[idx+1]-a[idx]);
	}
	a[idx]=val;
	 if(idx>0) {
		total+=Math.max(0,a[idx]-a[idx-1]);
	}
	 if(idx+1<a.length) {
		total+=Math.max(0,a[idx+1]-a[idx]);
	}
}
static int[] compreessarray(int c[]) {
	Stack<Integer> st=new Stack<>();
	for (int i = 0; i < c.length; i++) {
		if(st.isEmpty()||c[i]!=st.peek())
		st.add(c[i]);
	}
	b=new int[st.size()];
	for (int i = b.length-1; i>=0; i--) {
		b[i]=st.pop();
	}
	return b;
}
static int div4[];
static long [] reversearray(long a[]) {
	for (int i = 0; i <a.length/2; i++) {
		long temp=a[i];
		a[i]=a[a.length-i-1];
		a[a.length-i-1]=temp;
	}
	return a;
}
static int [] reversearray(int a[]) {
	for (int i = 0; i <=a.length/2; i++) {
		int temp=a[i];
		a[i]=a[a.length-i-1];
		a[a.length-i-1]=temp;
	}
	return a;
}
static class SegmentTree {		// 1-based DS, OOP
	
	int N; 			//the number of elements in the array as a power of 2 (i.e. after padding)
	int[] array, sTree, lazy;
	
	SegmentTree(int[] in)		
	{
		array = in; N = in.length - 1;
		sTree = new int[N<<1];		//no. of nodes = 2*N - 1, we add one to cross out index zero
		lazy = new int[N<<1];
		build(1,1,N);
	}
	
	void build(int node, int b, int e)	// O(n)
	{
		if(b == e)					
			sTree[node] = array[b];
		else						
		{
			int mid = b + e >> 1;
			build(node<<1,b,mid);
			build(node<<1|1,mid+1,e);
			sTree[node] = sTree[node<<1]+sTree[node<<1|1];
		}
	}
	
	
	void update_point(int index, int val)			// O(log n)
	{
		index += N - 1;				
		sTree[index] += val;			
		while(index>1)				
		{
			index >>= 1;
			sTree[index] = sTree[index<<1] + sTree[index<<1|1];		
		}
	}
	
	
	void update_range(int i, int j, int val)		// O(log n) 
	{
		update_range(1,1,N,i,j,val);
	}
	
	void update_range(int node, int b, int e, int i, int j, int val)
	{
		if(i > e || j < b)		
			return;
		if(b >= i && e <= j)		
		{
			sTree[node] += (e-b+1)*val;			
			lazy[node] += val;				
		}							
		else		
		{
			int mid = b + e >> 1;
			propagate(node, b, mid, e);
			update_range(node<<1,b,mid,i,j,val);
			update_range(node<<1|1,mid+1,e,i,j,val);
			sTree[node] = sTree[node<<1] + sTree[node<<1|1];		
		}
		
	}
	void propagate(int node, int b, int mid, int e)		
	{
		lazy[node<<1] += lazy[node];
		lazy[node<<1|1] += lazy[node];
		sTree[node<<1] += (mid-b+1)*lazy[node];		
		sTree[node<<1|1] += (e-mid)*lazy[node];		
		lazy[node] = 0;
	}
	
	int query(int i, int j)
	{
		return query(1,1,N,i,j);
	}
	
	int query(int node, int b, int e, int i, int j)	// O(log n)
	{
		if(i>e || j <b)
			return 0;		
		if(b>= i && e <= j)
			return sTree[node];
		int mid = b + e >> 1;
		propagate(node, b, mid, e);
		int q1 = query(node<<1,b,mid,i,j);
		int q2 = query(node<<1|1,mid+1,e,i,j);
		return q1 + q2;	
				
	}
}
static  TreeSet<Integer> ts=new TreeSet();
static HashMap<Integer, Integer> compress(int a[]) {
		 ts = new TreeSet<>();
		HashMap<Integer, Integer> hm = new HashMap<>();
		for (int x : a)
			ts.add(x);
		for (int x : ts) {
			hm.put(x, hm.size() + 1);
		}
		return hm;
	}	

	static  class FenwickTree { // one-based DS

		int n;
		int[] ft;
		
		FenwickTree(int size) { n = size; ft = new int[n+1]; }
		
		int rsq(int b) //O(log n)
		{
			int sum = 0;
			while(b > 0) { sum += ft[b]; b -= b & -b;}		//min?
			return sum;
		}
		
		int rsq(int a, int b) { return rsq(b) - rsq(a-1); }	
		
		void point_update(int k, int val)	//O(log n), update = increment
		{
			while(k <= n) { ft[k] += val; k += k & -k; }		//min?
		}
	}
	static ArrayList<Integer> euler=new ArrayList<>();
	

	static ArrayList<Integer> arr;

	
static long total;
static TreeMap<Integer,Integer> map1;

		static int zz;
//static int dp(int idx,int left,int state) {
//if(idx>=k-((zz-left)*2)||idx+1==n) {
//	return 0;
//}
//if(memo[idx][left][state]!=-1) {
//	return memo[idx][left][state];
//}
//int ans=a[idx+1]+dp(idx+1,left,0);
//if(left>0&&state==0&&idx>0) {
//	ans=Math.max(ans,a[idx-1]+dp(idx-1,left-1,1));
//}
//return memo[idx][left][state]=ans;
//}21
static HashMap<Integer,Integer> map;
static int maxa=0;
static int ff=123123;

static long[][] memo;


	static long modmod=998244353;
	static int dx[]= {1,-1,0,0};
	 static int dy[]= {0,0,1,-1};	
	static class BBOOK implements Comparable<BBOOK>{
		int t;
		int alice;
		int bob;
		public BBOOK(int x,int y,int z) {
		t=x;
		alice=y;
		bob=z;
		}
		@Override
		public int compareTo(BBOOK o) {
			return this.t-o.t;
		}
	}
	private static long lcm(long a2, long b2) {
	return (a2*b2)/gcd(a2,b2);
	}
	static class Edge implements Comparable<Edge>
	{
		int node;long cost ; long time; 
		
		Edge(int a, long b,long c) { node = a;	cost = b; time=c; }
		
		public int compareTo(Edge e){ return Long.compare(time,e.time);	}
	}
	
        static void sieve(int N)	// O(N log log N) 
	{
		isComposite = new int[N+1];					
		isComposite[0] = isComposite[1] = 1;			// 0 indicates a prime number
		primes = new ArrayList<Integer>();
 
		for (int i = 2; i <= N; ++i) 					//can loop till i*i <= N if primes array is not needed O(N log log sqrt(N)) 
			if (isComposite[i] == 0) 					//can loop in 2 and odd integers for slightly better performance
			{
				primes.add(i);
				if(1l * i * i <= N)
					for (int j = i * i; j <= N; j += i)	// j = i * 2 will not affect performance too much, may alter in modified sieve
						isComposite[j] = 1;
			}   
	}
    static TreeSet<Integer> factors;
    static TreeSet<Integer> primeFactors(int N)		// O(sqrt(N) / ln sqrt(N))
	{
		 factors = new TreeSet<Integer>();		//take abs(N) in case of -ve integers
		int idx = 0, p = primes.get(idx);

		while(1l*p * p <= N)
		{
			while(N % p == 0) { factors.add(p); N /= p; }
			p = primes.get(++idx);
		}

		if(N != 1)						// last prime factor may be > sqrt(N)
			factors.add(N);				// for integers whose largest prime factor has a power of 1
		return factors;
	}

	static class UnionFind {
		int[] p, rank, setSize;
		int numSets;
		int max[];

		public UnionFind(int N) {
			p = new int[numSets = N];
			rank = new int[N];
			setSize = new int[N];
			for (int i = 0; i < N; i++) {
				p[i] = i;
				setSize[i] = 1;
			}
		}

		public int findSet(int i) {
			return p[i] == i ? i : (p[i] = findSet(p[i]));
		}

		public boolean isSameSet(int i, int j) {
			return findSet(i) == findSet(j);
		}
		public int chunion(int i,int j, int x2) {
			if (isSameSet(i, j))
				return 0;
			numSets--;
			int x = findSet(i), y = findSet(j);
			int z=findSet(x2);
			p[x]=z;;
			p[y]=z;
		return x;
		}

		public void unionSet(int i, int j) {
			if (isSameSet(i, j))
				return;
			numSets--;
			int x = findSet(i), y = findSet(j);
			if (rank[x] > rank[y]) {
				p[y] = x;
				setSize[x] += setSize[y];


			} else {
				p[x] = y;
				setSize[y] += setSize[x];
				if (rank[x] == rank[y])
					rank[y]++;


			}
		}

		

		public int numDisjointSets() {
			return numSets;
		}

		public int sizeOfSet(int i) {
			return setSize[findSet(i)];
		}
	}


	static class Quad implements Comparable<Quad> {
		int u;
		int v;
		char state;
		int turns;

		public Quad(int i, int j, char c, int k) {
			u = i;
			v = j;
			state = c;
			turns = k;
		}

		public int compareTo(Quad e) {
			return (int) (turns - e.turns);
		}

	}

	static long manhatandistance(long x, long x2, long y, long y2) {
		return Math.abs(x - x2) + Math.abs(y - y2);
	}

	static long fib[];

	static long fib(int n) {
		if (n == 1 || n == 0) {
			return 1;
		}
		if (fib[n] != -1) {
			return fib[n];
		} else
			return fib[n] = ((fib(n - 2) % mod + fib(n - 1) % mod) % mod);
	}

	static class Point  implements Comparable<Point>{
		long x, y;

		Point(long counth, long counts) {
			x = counth;
			y = counts;
		}

		@Override
		public int compareTo(Point p )
		{
			return Long.compare(p.y*1l*x, p.x*1l*y);
		}
	}

	static TreeSet<Long> primeFactors(long N) // O(sqrt(N) / ln sqrt(N))
	{
		TreeSet<Long> factors = new TreeSet<Long>(); // take abs(N) in case of -ve integers
		int idx = 0, p = primes.get(idx);

		while (p * p <= N) {
			while (N % p == 0) {
				factors.add((long) p);
				N /= p;
			}
			if (primes.size() > idx + 1)
				p = primes.get(++idx);
			else
				break;
		}

		if (N != 1) // last prime factor may be > sqrt(N)
			factors.add(N); // for integers whose largest prime factor has a power of 1
		return factors;
	}

	static boolean visited[];

	/**
	 * static int bfs(int s) { Queue<Integer> q = new LinkedList<Integer>();
	 * q.add(s); int count=0; int maxcost=0; int dist[]=new int[n]; dist[s]=0;
	 * while(!q.isEmpty()) {
	 * 
	 * int u = q.remove(); if(dist[u]==k) { break; } for(Pair v: adj[u]) {
	 * maxcost=Math.max(maxcost, v.cost);
	 * 
	 * 
	 * 
	 * if(!visited[v.v]) {
	 * 
	 * visited[v.v]=true; q.add(v.v); dist[v.v]=dist[u]+1; maxcost=Math.max(maxcost,
	 * v.cost); } }
	 * 
	 * } return maxcost; }
	 **/

	static boolean[] vis2;

	static boolean f2 = false;

	static long[][] matMul(long[][] a2, long[][] b, int p, int q, int r) // C(p x r) = A(p x q) x (q x r) -- O(p x q x
																			// r)
	{
		long[][] C = new long[p][r];
		for (int i = 0; i < p; ++i) {
			for (int j = 0; j < r; ++j) {
				for (int k = 0; k < q; ++k) {
					C[i][j] = (C[i][j] + (a2[i][k] % mod * b[k][j] % mod)) % mod;
					C[i][j] %= mod;
				}

			}
		}
		return C;
	}
	public static int[] schuffle(int[] a2) {
		for (int i = 0; i < a2.length; i++) {
			int x =  (int) (Math.random() * a2.length);
			int temp = a2[x];
			a2[x] = a2[i];
			a2[i] = temp;
		}
		return a2;
	}


	static boolean vis[];
	static HashSet<Integer> set = new HashSet<Integer>();

	static long modPow(long ways, long count, long mod) // O(log e)
	{
		ways %= mod;
		long res = 1;
		while (count > 0) {
			if ((count & 1) == 1)
				res = (res * ways) % mod;
			ways = (ways * ways) % mod;
			count >>= 1;
		}
		return res % mod;
	}

	static long gcd(long l, long o) {
		if (o == 0) {
			return l;
		}
		return gcd(o, l % o);
	}

	static int[] isComposite;
	static int[] valid;

	static ArrayList<Integer> primes;
	static ArrayList<Integer> l1;

	
	static TreeSet<Integer> primus = new TreeSet<Integer>();
	
	static void sieveLinear(int N)
	{
		int[] lp = new int[N + 1];								//lp[i] = least prime divisor of i
		for(int i = 2; i <= N; ++i)
		{
			if(lp[i] == 0)
			{
				primus.add(i);
				lp[i] = i;
			}
			int curLP = lp[i];
			for(int p: primus)
				if(p > curLP || p * i > N)
					break;
				else
					lp[p * i] = i;
		}
	}
	

	public static long[] schuffle(long[] a2) {
		for (int i = 0; i < a2.length; i++) {
			int x =  (int) (Math.random() * a2.length);
			long temp = a2[x];
			a2[x] = a2[i];
			a2[i] = temp;
		}
		return a2;
	}

	static class Scanner {
		StringTokenizer st;
		BufferedReader br;

		public Scanner(InputStream system) {
			br = new BufferedReader(new InputStreamReader(system));
		}

		public Scanner(String file) throws Exception {
			br = new BufferedReader(new FileReader(file));
		}

		public String next() throws IOException {
			while (st == null || !st.hasMoreTokens())
				st = new StringTokenizer(br.readLine());
			return st.nextToken();
		}

		public String nextLine() throws IOException {
			return br.readLine();
		}

		public int nextInt() throws IOException {
			return Integer.parseInt(next());
		}

		public double nextDouble() throws IOException {
			return Double.parseDouble(next());
		}

		public char nextChar() throws IOException {
			return next().charAt(0);
		}

		public Long nextLong() throws IOException {
			return Long.parseLong(next());
		}

		public boolean ready() throws IOException {
			return br.ready();
		}

		public void waitForInput() throws InterruptedException {
			Thread.sleep(3000);
		}

		public int[] nxtArr(int n) throws IOException {
			int[] ans = new int[n];
			for (int i = 0; i < n; i++)
				ans[i] = nextInt();
			return ans;
		}
		
	}
	public static int[] sortarray(int a[]) {
		schuffle(a);
		Arrays.sort(a);
		return a;
		}
		public static long[] sortarray(long a[]) {
			schuffle(a);
			Arrays.sort(a);
			return a;
		}
		public static int[] readarray(int n) throws IOException {
			int a[]=new int[n];
			for (int i = 0; i < a.length; i++) {
			a[i]=sc.nextInt();
			}
			return a;
		}
		public static long[] readlarray(int n) throws IOException {
			long a[]=new long[n];
			for (int i = 0; i < a.length; i++) {
			a[i]=sc.nextLong();
			}
			return a;
		}
}