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 CPP):

Solution:
#include "bits/stdc++.h"
using namespace std;

const int N = 2e5 + 20;

int n, q, t, x;
set <pair <int, int>> row, col;

int main()
{
    scanf("%d%d", &n, &q);

    row.insert({n, n});
    col.insert({n, n});

    long long ans = 1ll * (n - 2) * (n - 2);
    while(q--)
    {
        scanf("%d%d", &t, &x);

        if(t == 1)
        {
            auto it = col.lower_bound({x, 0});
            ans -= (*it).second - 2;
            row.insert({(*it).second - 1, x});
        }
        else
        {
            auto it = row.lower_bound({x, 0});
            ans -= (*it).second - 2;
            col.insert({(*it).second - 1, x});
        }
    }

    printf("%lld\n", ans);
}