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:
// Utilities
import java.io.*;
import java.util.*;

public class Main {
	static int T;
	static int N, Q;
	static int[] a;
	static long[][][] st; // [i][0 = start with up, 1 = start with down][0 = end with up, 1 = end with down]
	static int l, r, tmp;
	
	public static void main(String[] args) throws IOException {
		T = in.iscan();
		while (T-- > 0) {
			N = in.iscan(); Q = in.iscan(); a = new int[N+1];
			for (int i = 1; i <= N; i++) a[i] = in.iscan();
			st = new long[4*N+5][2][2];
			init(1, 1, N);
			out.println(Math.max(Math.max(st[1][0][0],  st[1][0][1]), Math.max(st[1][1][0], st[1][1][1])));
			for (int i = 0; i < Q; i++) {
				l = in.iscan(); r = in.iscan();
				tmp = a[l];
				update(1, 1, N, l, a[r]);
				a[l] = a[r];
				update(1, 1, N, r, tmp);
				a[r] = tmp;
				out.println(Math.max(Math.max(st[1][0][0],  st[1][0][1]), Math.max(st[1][1][0], st[1][1][1])));
			}
		}
		out.close();
	} 
	
	static void init(int i, int l, int r) {
		if (l == r) {
			st[i][0][0] = a[l]; st[i][1][0] = 0;
			st[i][0][1] = 0; st[i][1][1] = -a[l];
			return;
		}
		int mid = (l+r)/2;
		init(2*i, l, mid);
		init(2*i+1, mid+1, r);
		st[i][0][0] = Math.max(st[2*i][0][1] + st[2*i+1][0][0], st[2*i][0][0] + st[2*i+1][1][0]);
		st[i][0][1] = Math.max(st[2*i][0][1] + st[2*i+1][0][1], st[2*i][0][0] + st[2*i+1][1][1]);
		st[i][1][0] = Math.max(st[2*i][1][0] + st[2*i+1][1][0], st[2*i][1][1] + st[2*i+1][0][0]);
		st[i][1][1] = Math.max(st[2*i][1][0] + st[2*i+1][1][1], st[2*i][1][1] + st[2*i+1][0][1]);
	}
	
	static void update(int i, int l, int r, int x, int v) {
		if (l == r) {
			st[i][0][0] = v; st[i][1][0] = 0;
			st[i][0][1] = 0; st[i][1][1] = -v;
			return;
		}
		int mid = (l+r)/2;
		if (x <= mid) update(2*i, l, mid, x, v);
		else update(2*i+1, mid+1, r, x, v);
		st[i][0][0] = Math.max(st[2*i][0][1] + st[2*i+1][0][0], st[2*i][0][0] + st[2*i+1][1][0]);
		st[i][0][1] = Math.max(st[2*i][0][1] + st[2*i+1][0][1], st[2*i][0][0] + st[2*i+1][1][1]);
		st[i][1][0] = Math.max(st[2*i][1][0] + st[2*i+1][1][0], st[2*i][1][1] + st[2*i+1][0][0]);
		st[i][1][1] = Math.max(st[2*i][1][0] + st[2*i+1][1][1], st[2*i][1][1] + st[2*i+1][0][1]);
	}
	
	static INPUT in = new INPUT(System.in);
	static PrintWriter out = new PrintWriter(System.out);
	private static class INPUT {

		private InputStream stream;
		private byte[] buf = new byte[1024];
		private int curChar, numChars;

		public INPUT (InputStream stream) {
			this.stream = stream;
		}

		public INPUT (String file) throws IOException {
			this.stream = new FileInputStream (file);
		}

		public int cscan () throws IOException {
			if (curChar >= numChars) {
				curChar = 0;
				numChars = stream.read (buf);
			}
			
			if (numChars == -1)
				return numChars;

			return buf[curChar++];
		}

		public int iscan () throws IOException {
			int c = cscan (), sgn = 1;
			
			while (space (c))
				c = cscan ();

			if (c == '-') {
				sgn = -1;
				c = cscan ();
			}

			int res = 0;

			do {
				res = (res << 1) + (res << 3);
				res += c - '0';
				c = cscan ();
			}
			while (!space (c));

			return res * sgn;
		}

		public String sscan () throws IOException {
			int c = cscan ();
			
			while (space (c))
				c = cscan ();

			StringBuilder res = new StringBuilder ();

			do {
				res.appendCodePoint (c);
				c = cscan ();
			}
			while (!space (c));

			return res.toString ();
		}

		public double dscan () throws IOException {
			int c = cscan (), sgn = 1;
			
			while (space (c))
				c = cscan ();

			if (c == '-') {
				sgn = -1;
				c = cscan ();
			}

			double res = 0;

			while (!space (c) && c != '.') {
				if (c == 'e' || c == 'E')
					return res * UTILITIES.fast_pow (10, iscan ());
				
				res *= 10;
				res += c - '0';
				c = cscan ();
			}

			if (c == '.') {
				c = cscan ();
				double m = 1;

				while (!space (c)) {
					if (c == 'e' || c == 'E')
						return res * UTILITIES.fast_pow (10, iscan ());

					m /= 10;
					res += (c - '0') * m;
					c = cscan ();
				}
			}

			return res * sgn;
		}

		public long lscan () throws IOException {
			int c = cscan (), sgn = 1;
			
			while (space (c))
				c = cscan ();

			if (c == '-') {
				sgn = -1;
				c = cscan ();
			}

			long res = 0;

			do {
				res = (res << 1) + (res << 3);
				res += c - '0';
				c = cscan ();
			}
			while (!space (c));

			return res * sgn;
		}

		public boolean space (int c) {
			return c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == -1;
		}
	}

	public static class UTILITIES {

		static final double EPS = 10e-6;

		public static int lower_bound (int[] arr, int x) {
			int low = 0, high = arr.length, mid = -1;

			while (low < high) {
				mid = (low + high) / 2;

				if (arr[mid] >= x)
					high = mid;
				else
					low = mid + 1;
			}

			return low;
		}

		public static int upper_bound (int[] arr, int x) {
			int low = 0, high = arr.length, mid = -1;

			while (low < high) {
				mid = (low + high) / 2;

				if (arr[mid] > x)
					high = mid;
				else
					low = mid + 1;
			}

			return low;
		}

		public static int gcd (int a, int b) {
			return b == 0 ? a : gcd (b, a % b);
		}

		public static int lcm (int a, int b) {
			return a * b / gcd (a, b);
		}

		public static long fast_pow_mod (long b, long x, int mod) {
			if (x == 0) return 1;
			if (x == 1) return b;
			if (x % 2 == 0) return fast_pow_mod (b * b % mod, x / 2, mod) % mod;

			return b * fast_pow_mod (b * b % mod, x / 2, mod) % mod;
		}

		public static int fast_pow (int b, int x) {
			if (x == 0) return 1;
			if (x == 1) return b;
			if (x % 2 == 0) return fast_pow (b * b, x / 2);

			return b * fast_pow (b * b, x / 2);
		}

		public static long choose (long n, long k) {
			k = Math.min (k, n - k);
			long val = 1;

			for (int i = 0; i < k; ++i)
				val = val * (n - i) / (i + 1);

			return val;
		}

		public static long permute (int n, int k) {
			if (n < k) return 0;
			long val = 1;

			for (int i = 0; i < k; ++i)
				val = (val * (n - i));

			return val;
		}
	}
}