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 PYTHON3):

Solution:
#### IMPORTANT LIBRARY ####
 
############################
### DO NOT USE import random --> 250ms to load the library
############################
 
### In case of extra libraries: https://github.com/cheran-senthil/PyRival
 
######################
####### IMPORT #######
######################
 
from functools import cmp_to_key 
from collections import deque
from heapq import heappush, heappop
from math import log, ceil
 
######################
#### STANDARD I/O ####
######################
 
import sys
import os
from io import BytesIO, IOBase
 
BUFSIZE = 8192
 
class FastIO(IOBase):
    newlines = 0
    def __init__(self, file):
        self._fd = file.fileno()
        self.buffer = BytesIO()
        self.writable = "x" in file.mode or "r" not in file.mode
        self.write = self.buffer.write if self.writable else None
 
    def read(self):
        while True:
            b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
            if not b:
                break
            ptr = self.buffer.tell()
            self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
        self.newlines = 0
        return self.buffer.read()
 
    def readline(self):
        while self.newlines == 0:
            b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
            self.newlines = b.count(b"\n") + (not b)
            ptr = self.buffer.tell()
            self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
        self.newlines -= 1
        return self.buffer.readline()
 
    def flush(self):
        if self.writable:
            os.write(self._fd, self.buffer.getvalue())
            self.buffer.truncate(0), self.buffer.seek(0)
 
class IOWrapper(IOBase):
    def __init__(self, file):
        self.buffer = FastIO(file)
        self.flush = self.buffer.flush
        self.writable = self.buffer.writable
        self.write = lambda s: self.buffer.write(s.encode("ascii"))
        self.read = lambda: self.buffer.read().decode("ascii")
        self.readline = lambda: self.buffer.readline().decode("ascii")
 
if sys.version_info[0] < 3:
    sys.stdin, sys.stdout = FastIO(sys.stdin), FastIO(sys.stdout)
else:
    sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
 
def print(*args, **kwargs):
    sep, file = kwargs.pop("sep", " "), kwargs.pop("file", sys.stdout)
    at_start = True
    for x in args:
        if not at_start:
            file.write(sep)
        file.write(str(x))
        at_start = False
    file.write(kwargs.pop("end", "\n"))
    if kwargs.pop("flush", False):
        file.flush()
 
def inp():
    return sys.stdin.readline().rstrip("\r\n")  # for fast input
    
def ii():
    return int(inp())
 
def li(lag = 0):
    l = list(map(int, inp().split()))
    if lag != 0:
        for i in range(len(l)):
            l[i] += lag
    return l
    
def mi(lag = 0):
    matrix = list()
    for i in range(n):
        matrix.append(li(lag))
    return matrix
 
def sli(): #string list
    return list(map(str, inp().split()))
 
def print_list(lista, space = " "):
    print(space.join(map(str, lista)))
 
######################
##### UNION FIND #####
######################
 
class UnionFind:
    def __init__(self, n):
        self.parent = list(range(n))
        self.size = [1] * n
        self.num_sets = n
 
    def find(self, a):
        to_update = []
        while a != self.parent[a]:
            to_update.append(a)
            a = self.parent[a]
        for b in to_update:
            self.parent[b] = a
        return self.parent[a]
 
    def merge(self, a, b):
        a = self.find(a)
        b = self.find(b)
        if a == b:
            return
        if self.size[a] < self.size[b]:
            a, b = b, a
        self.num_sets -= 1
        self.parent[b] = a
        self.size[a] += self.size[b]
 
    def set_size(self, a):
        return self.size[self.find(a)]
 
    def __len__(self):
        return self.num_sets
 
######################
### BISECT METHODS ###
######################
 
def bisect_left(a, x):
    """i tale che a[i] >= x e a[i-1] < x"""
    left = 0
    right = len(a)
    while left < right:
        mid = (left+right)//2
        if a[mid] < x: 
            left = mid+1
        else: 
            right = mid
    return left
 
def bisect_right(a, x):
    """i tale che a[i] > x e a[i-1] <= x"""
    left = 0
    right = len(a)
    while left < right:
        mid = (left+right)//2
        if a[mid] > x: 
            right = mid
        else: 
            left = mid+1
    return left
 
def bisect_elements(a, x):
    """elementi pari a x nell'árray sortato"""
    return bisect_right(a, x) - bisect_left(a, x)
 
######################
#### CUSTOM SORT #####
######################
 
def custom_sort(lista):
    def cmp(x,y): 
        if x+y>y+x:
            return 1 
        else:
            return -1
    return sorted(lista, key = cmp_to_key(cmp))
 
######################
### MOD OPERATION ####
######################
 
MOD = 10**9 + 7
maxN = 10**5
FACT = [0] * maxN
 
def add(x, y):
    return (x+y) % MOD
   
def multiply(x, y):
    return (x*y) % MOD
 
def power(x, y):
    if y == 0:
        return 1
    elif y % 2:
        return multiply(x, power(x, y-1))
    else:
        a = power(x, y//2)
        return multiply(a, a)
 
def inverse(x):
    return power(x, MOD-2)
   
def divide(x, y):
    return multiply(x, inverse(y))
 
def allFactorials():
    FACT[0] = 1
    for i in range(1, maxN):
        FACT[i] = multiply(i, FACT[i-1])
 
def coeffBinom(n, k):
    if n < k:
        return 0
    return divide(FACT[n], multiply(FACT[k], FACT[n-k]))
 
######################
#### GCD & PRIMES ####
######################
 
def primes(N):
    smallest_prime = [1] * (N+1)
    prime = [] 
    smallest_prime[0] = 0
    smallest_prime[1] = 0
    for i in range(2, N+1):  
        if smallest_prime[i] == 1: 
            prime.append(i)
            smallest_prime[i] = i
        j = 0
        while (j < len(prime) and i * prime[j] <= N): 
            smallest_prime[i * prime[j]] = min(prime[j], smallest_prime[i])
            j += 1
    return prime, smallest_prime
 
def gcd(a, b):
    a = abs(a)
    b = abs(b)
    s, t, r = 0, 1, b
    old_s, old_t, old_r = 1, 0, a
    while r != 0:
        quotient = old_r//r
        old_r, r = r, old_r - quotient*r
        old_s, s = s, old_s - quotient*s
        old_t, t = t, old_t - quotient*t
    return old_r, old_s, old_t #gcd, x, y for ax+by=gcd
 
######################
#### GRAPH ALGOS #####
######################
 
# ZERO BASED GRAPH
def create_graph(n, m, undirected = 1, unweighted = 1):
    graph = [[] for i in range(n)]
    if unweighted:
        for i in range(m):
            [x, y] = li(lag = -1)
            graph[x].append(y)
            if undirected:
                graph[y].append(x)
    else:
        for i in range(m):
            [x, y, w] = li(lag = -1)
            w += 1
            graph[x].append([y,w])
            if undirected:
                graph[y].append([x,w])
    return graph
 
def create_tree(n, unweighted = 1):
    children = [[] for i in range(n)]
    if unweighted:
        for i in range(n-1):
            [x, y] = li(lag = -1)
            children[x].append(y)
            children[y].append(x)
    else:
        for i in range(n-1):
            [x, y, w] = li(lag = -1)
            w += 1
            children[x].append([y, w])
            children[y].append([x, w])
    return children
   
def create_edges(m, unweighted = 0):
    edges = list()
    if unweighted:
        for i in range(m):
            edges.append(li(lag = -1))
    else:
        for i in range(m):
            [x, y, w] = li(lag = -1)
            w += 1
            edges.append([w,x,y])
    return edges
 
def dist(tree, n, A, B = -1):
    s = [[A, 0]]
    massimo, massimo_nodo = 0, 0
    distanza = -1
    v = [-1] * n
    while s:
        el, dis = s.pop() 
        if dis > massimo:
            massimo = dis
            massimo_nodo = el
        if el == B:
            distanza = dis
        for child in tree[el]:
            if v[child] == -1:
                v[child] = 1
                s.append([child, dis+1])
    return massimo, massimo_nodo, distanza
 
def diameter(tree):
    _, foglia, _ = dist(tree, n, 0)
    diam, _, _ = dist(tree, n, foglia)
    return diam
   
def dfs(graph, n, A):
    v = [-1] * n
    s = [[A, 0]]
    v[A] = 0
    while s:
        el, dis = s.pop()
        for child in graph[el]:
            if v[child] == -1:
                v[child] = dis + 1
                s.append([child, dis + 1])
    return v #visited: -1 if not visited, otherwise v[B] is the distance in terms of edges
 
def bfs(graph, n, A):
    v = [-1] * n
    s = deque()
    s.append([A, 0])
    v[A] = 0
    while s:
        el, dis = s.popleft()
        for child in graph[el]:
            if v[child] == -1:
                v[child] = dis + 1
                s.append([child, dis + 1])
    return v #visited: -1 if not visited, otherwise v[B] is the distance in terms of edges
 
def connected(graph, n):
    v = dfs(graph, n, 0)
    for el in v:
        if el == -1:
            return False
    return True
 
# NON DIMENTICARTI DI PRENDERE GRAPH COME DIRETTO
def topological(graph, n):
    indegree = [0] * n
    for el in range(n):
        for child in graph[el]:
            indegree[child] += 1
    s = deque()
    for el in range(n):
        if indegree[el] == 0:
            s.append(el)
    order = []
    while s:
        el = s.popleft()
        order.append(el)
        for child in graph[el]:
            indegree[child] -= 1
            if indegree[child] == 0:
                s.append(child)
    if n == len(order):
        return False, order #False == no cycle
    else:
        return True, [] #True == there is a cycle and order is useless
 
# ASSUMING CONNECTED
def bipartite(graph, n):
    color = [-1] * n
    color[0] = 0
    s = [0]
    while s:
        el = s.pop()
        for child in graph[el]:
            if color[child] == color[el]:
                return False
            if color[child] == -1:
                s.append(child)
            color[child] = 1 - color[el]
    return True
 
# SHOULD BE DIRECTED AND WEIGHTED
def dijkstra(graph, n, A):
    dist = [float('inf') for i in range(n)]
    prev = [-1 for i in range(n)]
    dist[A] = 0
    pq = []
    heappush(pq, [0, A])
    while pq:
        [d_v, v] = heappop(pq)
        if (d_v != dist[v]):
            continue
        for to, w in graph[v]:
            if dist[v] + w < dist[to]:
                dist[to] = dist[v] + w
                prev[to] = v
                heappush(pq, [dist[to], to])
    return dist, prev
 
# SHOULD BE DIRECTED AND WEIGHTED
def dijkstra_0_1(graph, n, A):
    dist = [float('inf') for i in range(n)]
    dist[A] = 0
    p = deque()
    p.append(A)
    while p:
        v = p.popleft()
        for to, w in graph[v]:
            if dist[v] + w < dist[to]:
                dist[to] = dist[v] + w
            if w == 1:
                q.append(to)
            else:
                q.appendleft(to)
    return dist
 
#SHOULD BE WEIGHTED (AND UNDIRECTED)
def floyd_warshall(graph, n):
    dist = [[float('inf') for _ in range(n)] for _ in range(n)]
    for i in range(n):
        dist[i][i] = 0
        for child, d in graph[i]:
            dist[i][child] = d
            dist[child][i] = d
    for k in range(n):
        for i in range(n):
            for j in range(j):
                dist[i][j] = min(dist[i][j], dist[i][k] + dist[k][j])
    return dist
 
#EDGES [w,x,y]
def minimum_spanning_tree(edges, n):
    edges = sorted(edges)
    union_find = UnionFind(n) #implemented above
    used_edges = list()
    for w, x, y in edges:
        if union_find.find(x) != union_find.find(y):
            union_find.merge(x, y)
            used_edges.append([w,x,y])
    return used_edges
 
#FROM A GIVEN ROOT, RECOVER THE STRUCTURE
def parents_children_root_unrooted_tree(tree, n, root = 0):
    q = deque()
    visited = [0] * n
    parent = [-1] * n
    children = [[] for i in range(n)]
    q.append(root)
    while q:
        all_done = 1
        visited[q[0]] = 1
        for child in tree[q[0]]:
            if not visited[child]:
                all_done = 0
                q.appendleft(child)
        if all_done:
            for child in tree[q[0]]:   
                if parent[child] == -1:
                    parent[q[0]] = child
                    children[child].append(q[0])
            q.popleft()
    return parent, children
 
# CALCULATING LONGEST PATH FOR ALL THE NODES 
def all_longest_path_passing_from_node(parent, children, n):
    q = deque()
    visited = [len(children[i]) for i in range(n)]
    downwards = [[0,0] for i in range(n)]
    upward = [1] * n
    longest_path = [1] * n
    for i in range(n):
        if not visited[i]:
            q.append(i)
            downwards[i] = [1,0]
    while q:
        node = q.popleft()
        if parent[node] != -1:
            visited[parent[node]] -= 1
            if not visited[parent[node]]:
                q.append(parent[node])
        else:
            root = node
        for child in children[node]:   
            downwards[node] = sorted([downwards[node][0], downwards[node][1], downwards[child][0] + 1], reverse = True)[0:2]
    s = [node]
    while s:
        node = s.pop()
        if parent[node] != -1:
            if downwards[parent[node]][0] == downwards[node][0] + 1:
                upward[node] = 1 + max(upward[parent[node]], downwards[parent[node]][1])
            else:
                upward[node] = 1 + max(upward[parent[node]], downwards[parent[node]][0])
        longest_path[node] = downwards[node][0] + downwards[node][1] + upward[node] - min([downwards[node][0], downwards[node][1], upward[node]]) - 1
        for child in children[node]:
            s.append(child)
    return longest_path
 
def finding_ancestors(parent, queries, n):
    steps = int(ceil(log(n, 2)))
    ancestors = [[-1 for i in range(n)] for j in range(steps)]
    ancestors[0] = parent
    for i in range(1, steps):
        for node in range(n):
            if ancestors[i-1][node] != -1:
                ancestors[i][node] = ancestors[i-1][ancestors[i-1][node]]
    result = []
    for node, k in queries:
        ans = node
        if k >= n:
            ans = -1
        i = 0
        while k > 0 and ans != -1:
            if k % 2:
                ans = ancestors[i][ans]
            k = k // 2
            i += 1
        result.append(ans)
    return result #Preprocessing in O(n log n). For each query O(log k)
 
### TBD SUCCESSOR GRAPH 7.5
### TBD TREE QUERIES 10.2 da 2 a 4
### TBD ADVANCED TREE 10.3
### TBD GRAPHS AND MATRICES 11.3.3 e 11.4.3 e 11.5.3 (ON GAMES)
 
######################
####### OTHERS #######
######################
 
def prefix_sum(arr):
    r = [0] * (len(arr)+1)
    for i, el in enumerate(arr):
        r[i+1] = r[i] + el
    return r
   
def nearest_from_the_left_smaller_elements(arr):
    n = len(arr)
    res = [-1] * n
    s = []
    for i, el in enumerate(arr):
        while s and s[-1] >= el:
            s.pop()
        if s:
            res[i] = s[-1]
        s.append(el)
    return res
 
def sliding_window_minimum(arr, k):
    res = []
    q = deque()
    for i, el in enumerate(arr):
        while q and arr[q[-1]] >= el:
            q.pop()
        q.append(i)
        while q and q[0] <= i - k:
            q.popleft()
        if i >= k-1:
            res.append(arr[q[0]])
    return res
 
### TBD COUNT ELEMENT SMALLER THAN SELF
 
######################
## END OF LIBRARIES ##
######################

for test in range(ii()):
    n, q = li()
    a = li()
    neg = [-float('inf')] * n
    pos = [-float('inf')] * n
    neg[n-1] = -a[n-1]
    pos[n-1] = a[n-1]
    for i in range(n-2, -1, -1):
        pos[i] = max(a[i]+neg[i+1], pos[i+1], a[i])
        neg[i] = max(-a[i]+pos[i+1], neg[i+1], -a[i])
    print(max(pos))