Problem: There is a grid with N rows and N columns of squares. Let (i, j) be the square at the i-th row from the top and the j-th column from the left.

Each of the central (N-2) \times (N-2) squares in the grid has a black stone on it. Each of the 2N - 1 squares on the bottom side and the right side has a white stone on it.

Q queries are given. We ask you to process them in order. There are two kinds of queries. Their input format and description are as follows:

* `1 x`: Place a white stone on (1, x). After that, for each black stone between (1, x) and the first white stone you hit if you go down from (1, x), replace it with a white stone.
* `2 x`: Place a white stone on (x, 1). After that, for each black stone between (x, 1) and the first white stone you hit if you go right from (x, 1), replace it with a white stone.



How many black stones are there on the grid after processing all Q queries?

Constraints

* 3 \leq N \leq 2\times 10^5
* 0 \leq Q \leq \min(2N-4,2\times 10^5)
* 2 \leq x \leq N-1
* Queries are pairwise distinct.

Input

Input is given from Standard Input in the following format:


N Q
Query_1
\vdots
Query_Q


Output

Print how many black stones there are on the grid after processing all Q queries.

Examples

Input

5 5
1 3
2 3
1 4
2 2
1 2


Output

1


Input

200000 0


Output

39999200004


Input

176527 15
1 81279
2 22308
2 133061
1 80744
2 44603
1 170938
2 139754
2 15220
1 172794
1 159290
2 156968
1 56426
2 77429
1 97459
2 71282


Output

31159505795

Solution (in PYTHON3):

Solution:
N,Q = map(int,input().split())
Min_x = [-1 for _ in range(N)] #一番遠い位置にある黒(0index)
#Min_x[N-1] = -1
Min_y = [-1 for _ in range(N)] #一番遠い位置にある黒(0index)
#Min_y[N-1] = -1
black = (N-2)**2
a = N-2; b = N-2 #左上の四角の部分
for query in range(Q):
  #print(black)
  temp = list(map(int,input().split()))#1index #x-=1 #0index
  if temp[0] == 1: #上に置く
    w = temp[1]-1 #0index
    if w <= b:
      #左上の更新する四角の中の場合、更新前の値が裏返りに使われる。
      black -= a #一番遠い黒-1 (0行目にはない）裏返る。0indexなのでそのまま引いてよい。
      for yoko in range(w,b+1): #w~bまで
        Min_y[yoko] = a
      b = w-1
    else:
      #既に更新する四角から外れている場合には既存の値を入れる。
      black -= Min_y[w] #一番遠い黒-1 (0行目にはない）裏返る。0indexなのでそのまま引いてよい。
  else: #横に置く
    h = temp[1]-1 #0index
    if h <= a:
      #左上の更新する四角の中の場合、更新前の値が裏返りに使われる。
      black -= b #一番遠い黒-1 (0行目にはない）裏返る。0indexなのでそのまま引いてよい。
      #print(h,a)
      for tate in range(h,a+1): #w~bまで
        Min_x[tate] = b
      a = h-1
    else:
      #既に更新する四角から外れている場合には既存の値を入れる。
      black -= Min_x[h] #一番遠い黒-1 (0行目にはない）裏返る。0indexなのでそのまま引いてよい。      
  #print("x",Min_x)
  #print("y",Min_y)
  #print("black", black)
print(black)