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:
import sys
from operator import ge

def input(): return sys.stdin.readline().strip()
def list2d(a, b, c): return [[c for j in range(b)] for i in range(a)]
def list3d(a, b, c, d): return [[[d for k in range(c)] for j in range(b)] for i in range(a)]
def list4d(a, b, c, d, e): return [[[[e for l in range(d)] for k in range(c)] for j in range(b)] for i in range(a)]
def ceil(x, y=1): return int(-(-x // y))
def INT(): return int(input())
def MAP(): return map(int, input().split())
def LIST(N=None): return list(MAP()) if N is None else [INT() for i in range(N)]
def Yes(): print('Yes')
def No(): print('No')
def YES(): print('YES')
def NO(): print('NO')
sys.setrecursionlimit(10**9)
INF = 10**19
MOD = 10**9 + 7
EPS = 10**-10

class SegTree:
    """
    セグメント木
    1.update: i番目の値をxに更新する
    2.query: 区間[l, r)の値を得る
    """

    def __init__(self, n, func, intv, A=[]):
        """
        :param n: 要素数(0-indexed)
        :param func: 値の操作に使う関数(min, max, add, gcdなど)
        :param intv: 要素の初期値(単位元)
        :param A: 初期化に使うリスト(オプション)
        """
        self.n = n
        self.func = func
        self.intv = intv
        # nより大きい2の冪数
        n2 = 1
        while n2 < n:
            n2 <<= 1
        self.n2 = n2
        self.tree = [self.intv] * (n2 << 1)
        # 初期化の値が決まっている場合
        if A:
            # 1段目(最下段)の初期化
            for i in range(n):
                self.tree[n2+i] = A[i]
            # 2段目以降の初期化
            for i in range(n2-1, -1, -1):
                self.tree[i] = self.func(self.tree[i*2], self.tree[i*2+1])

    def update(self, i, x):
        """
        i番目の値をxに更新
        :param i: index(0-indexed)
        :param x: update value
        """
        i += self.n2
        self.tree[i] = x
        while i > 0:
            i >>= 1
            self.tree[i] = self.func(self.tree[i*2], self.tree[i*2+1])

    def add(self, i, x):
        self.update(i, self.get(i) + x)

    def query(self, a, b):
        """
        [a, b)の値を得る
        :param a: index(0-indexed)
        :param b: index(0-indexed)
        """
        l = a + self.n2
        r = b + self.n2
        s = self.intv
        while l < r:
            if r & 1:
                r -= 1
                s = self.func(s, self.tree[r])
            if l & 1:
                s = self.func(s, self.tree[l])
                l += 1
            l >>= 1
            r >>= 1
        return s

    def get(self, i):
        """ 一点取得 """
        return self.tree[i+self.n2]

    def all(self):
        """ 全区間[0, n)の取得 """
        return self.tree[1]

    def bisearch_fore(self, l, r, x, func):
        """ 区間[l,r]で左から最初にxに対して比較の条件を満たすような値が出現する位置 """

        ok = r + 1
        ng = l - 1
        while ng+1 < ok:
            mid = (ok+ng) // 2
            if func(self.query(l, mid+1), x):
                ok = mid
            else:
                ng = mid
        if ok != r + 1:
            return ok
        else:
            return INF

    def print(self):
        for i in range(self.n):
            print(self.get(i), end=' ')
        print()

N, Q = MAP()
ans = (N-2) * (N-2)
seg1 = SegTree(N+1, max, -INF, [0]*(N+1))
seg2 = SegTree(N+1, max, -INF, [0]*(N+1))
seg1.update(N, N)
seg2.update(N, N)
for _ in range(Q):
    t, x = MAP()
    if t == 1:
        res = seg1.bisearch_fore(0, N, x, ge)
        ans -= res - 2
        seg2.update(x, res)
    else:
        res = seg2.bisearch_fore(0, N, x, ge)
        ans -= res - 2
        seg1.update(x, res)
print(ans)