Problem: Vova decided to clean his room. The room can be represented as the coordinate axis OX. There are n piles of trash in the room, coordinate of the i-th pile is the integer p_i. All piles have different coordinates.

Let's define a total cleanup as the following process. The goal of this process is to collect all the piles in no more than two different x coordinates. To achieve this goal, Vova can do several (possibly, zero) moves. During one move, he can choose some x and move all piles from x to x+1 or x-1 using his broom. Note that he can't choose how many piles he will move.

Also, there are two types of queries:

  * 0 x — remove a pile of trash from the coordinate x. It is guaranteed that there is a pile in the coordinate x at this moment. 
  * 1 x — add a pile of trash to the coordinate x. It is guaranteed that there is no pile in the coordinate x at this moment. 



Note that it is possible that there are zero piles of trash in the room at some moment.

Vova wants to know the minimum number of moves he can spend if he wants to do a total cleanup before any queries. He also wants to know this number of moves after applying each query. Queries are applied in the given order. Note that the total cleanup doesn't actually happen and doesn't change the state of piles. It is only used to calculate the number of moves.

For better understanding, please read the Notes section below to see an explanation for the first example.

Input

The first line of the input contains two integers n and q (1 ≤ n, q ≤ 10^5) — the number of piles in the room before all queries and the number of queries, respectively.

The second line of the input contains n distinct integers p_1, p_2, ..., p_n (1 ≤ p_i ≤ 10^9), where p_i is the coordinate of the i-th pile.

The next q lines describe queries. The i-th query is described with two integers t_i and x_i (0 ≤ t_i ≤ 1; 1 ≤ x_i ≤ 10^9), where t_i is 0 if you need to remove a pile from the coordinate x_i and is 1 if you need to add a pile to the coordinate x_i. It is guaranteed that for t_i = 0 there is such pile in the current set of piles and for t_i = 1 there is no such pile in the current set of piles.

Output

Print q+1 integers: the minimum number of moves Vova needs to do a total cleanup before the first query and after each of q queries.

Examples

Input


5 6
1 2 6 8 10
1 4
1 9
0 6
0 10
1 100
1 50


Output


5
7
7
5
4
8
49


Input


5 8
5 1 2 4 3
0 1
0 2
0 3
0 4
0 5
1 1000000000
1 1
1 500000000


Output


3
2
1
0
0
0
0
0
499999999

Note

Consider the first example.

Initially, the set of piles is [1, 2, 6, 8, 10]. The answer before the first query is 5 because you can move all piles from 1 to 2 with one move, all piles from 10 to 8 with 2 moves and all piles from 6 to 8 with 2 moves.

After the first query, the set becomes [1, 2, 4, 6, 8, 10]. Then the answer is 7 because you can move all piles from 6 to 4 with 2 moves, all piles from 4 to 2 with 2 moves, all piles from 2 to 1 with 1 move and all piles from 10 to 8 with 2 moves.

After the second query, the set of piles becomes [1, 2, 4, 6, 8, 9, 10] and the answer is the same (and the previous sequence of moves can be applied to the current set of piles).

After the third query, the set of piles becomes [1, 2, 4, 8, 9, 10] and the answer is 5 because you can move all piles from 1 to 2 with 1 move, all piles from 2 to 4 with 2 moves, all piles from 10 to 9 with 1 move and all piles from 9 to 8 with 1 move.

After the fourth query, the set becomes [1, 2, 4, 8, 9] and the answer is almost the same (the previous sequence of moves can be applied without moving piles from 10).

After the fifth query, the set becomes [1, 2, 4, 8, 9, 100]. You can move all piles from 1 and further to 9 and keep 100 at its place. So the answer is 8.

After the sixth query, the set becomes [1, 2, 4, 8, 9, 50, 100]. The answer is 49 and can be obtained with almost the same sequence of moves as after the previous query. The only difference is that you need to move all piles from 50 to 9 too.

Solution (in PYTHON3):

Solution:
import os
import sys
from io import BytesIO, IOBase
 
from types import GeneratorType
BUFSIZE = 8192
class FastIO(IOBase):
    newlines = 0
 
    def __init__(self, file):
        import os
        self.os = os
        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 = self.os.read(self._fd, max(self.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 = self.os.read(self._fd, max(self.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:
            self.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")
 
 
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
 
import time
start_time = time.time()

import collections as col, math

def getInts():
    return [int(s) for s in input().split()]

def getInt():
    return int(input())

def getStrs():
    return [s for s in input().split()]

def getStr():
    return input()

def listStr():
    return list(input())

def solve():
    # function to build the tree  
    def build(arr) : 
      
        # insert leaf nodes in tree  
        for i in range(n) :
            x = arr[i]
            tree[n + i] = (counts[x],x,x,0);  
          
        # build the tree by calculating parents  
        for i in range(n - 1, 0, -1) :
            a, b = tree[i << 1], tree[i << 1 | 1]
            if a[0] == 0:
                tree[i] = b
            elif b[0] == 0:
                tree[i] = a
            else:
                if i << 1 == 2 and a[2] > b[2] and a[0] > 0:
                    #gap missing gap to find is the difference between the max of the right tree and the element above it
                    gap = tree2[1][1] - b[2]
                else:
                    gap = b[1]-a[2]
                tree[i] = (a[0]+b[0],min(a[1],b[1]),max(a[2],b[2]),max(a[3],b[3],gap));  
      
    # function to update a tree node  
    def updateTreeNode(p, value) :
        # set value at position p  
        counts[p] = value
        #print(points_index[p]+n,len(points),len(tree))
        #print(tree)
        tree[points_index[p] + n] = (value, p, p,0);
          
        # move upward and update parents  
        i = points_index[p]+n; 
          
        while i > 1 :
            #print(i)
            i >>= 1
            #print(i,tree)
            a, b = tree[i << 1], tree[i << 1 | 1]
            if a[0] == 0:
                tree[i] = b
            elif b[0] == 0:
                tree[i] = a
            else:
                #if a[2] > b[2]:
                #    gap = tree2[1][1] - b[2]
                #else:
                gap = b[1]-a[2]
                tree[i] = (a[0]+b[0],min(a[1],b[1]),max(a[2],b[2]),max(a[3],b[3],gap));
   
    n, Q = getInts()
    P = getInts()
    queries = []
    points = set(P)
    for q in range(Q):
        t, x = getInts()
        queries.append((t,x))
        points.add(x)
    counts = col.defaultdict(int)
    for p in P:
        counts[p] = 1
    points = sorted(list(points))
    points_index = col.defaultdict(int)
    n = len(points)
    two_pow = 1
    while two_pow <= 2*n:
        two_pow *= 2
    n = two_pow//2
    #add some dummy points
    j = 10**10
    while len(points) < n:
        points.append(j+x)
        j += 1
    tree = [0]*(two_pow)
    build(points)
    for i, point in enumerate(points):
        points_index[point] = i
    print(tree[1][2]-tree[1][1]-tree[1][3])
    for t, x in queries:
        updateTreeNode(x,t)

        print(tree[1][2]-tree[1][1]-tree[1][3])
    return
    #segment tree of counts
    #in each node, we store: the count (only ever 1 or 0), then minimum value, the maximum value
    #somehow need to come up with a way to store maxgap
    #the response to each query is max-min-maxgap
    

solve()
    
#print(time.time()-start_time)