题意：有一个M*M的网格，坐标[0...M-1,0...M-1] 网格里面有两个y坐标同样的宾馆A和B。以及n个餐厅，宾馆AB里面各有一个餐厅，编号1,2，其它餐厅编号3-n。如今你打算新开一家餐厅。须要考察一下可能的位置。一个位置p是“好位置”的条件是：当且仅当对于已有的每一个餐厅q。要么p比q离A近。要么p比q离B近。即dist(p,A) < dist(q,A) 或者 dist(p,B) < dist(q,B)  问“好位置”的个数

#include<cstdio>

#include<cstring>

#include<cmath>

#include<cstdlib>

#include<iostream>

#include<algorithm>

#include<vector>

#include<map>

#include<queue>

#include<stack>

#include<string>

#include<map>

#include<set>

#define eps 1e-6

#define LL long long

using namespace std;  

const int maxn = 50000 + 50;

const int maxm = 60000 + 50;

const int INF = 0x3f3f3f3f;

int borderl[maxm], borderr[maxm], rest[maxm];

int m, n, begx, endx, level;//begx。endx记录ab的横坐标 

void init() {

	memset(rest, -1, sizeof(rest));

	cin >> m >> n;

	int tx, ty;

	cin >> tx >> ty; rest[tx] = ty; begx = tx;

	cin >> tx >> ty; rest[tx] = ty; endx = tx;

	if(begx > endx) swap(begx, endx);

	level = ty;

	for(int i = 3; i <= n; i++) {

		cin >> tx >> ty;

		if(ty < level) ty = 2*level - ty;

		if(rest[tx] != -1) rest[tx] = min(rest[tx], ty);

		else rest[tx] = ty;

	}

}

void solve() {

	int ans = 0;

	borderl[begx] = borderl[endx] = level;

	borderr[begx] = borderr[endx] = level;

	for(int i = begx + 1; i < endx; i++) {

		if(rest[i] != -1) {

			borderl[i] = min(borderl[i-1]+1, rest[i]);

		}

		else {

			borderl[i] = borderl[i-1] + 1;

		}

	}

	for(int i = endx - 1; i > begx; i--) {

		if(rest[i] != -1) {

			borderr[i] = min(borderr[i+1]+1, rest[i]);

		}

		else {

			borderr[i] = borderr[i+1] + 1;

		}

	}

	for(int i = begx + 1; i < endx; i++)

		 ans += max(0, min(min(borderl[i], borderr[i]), m)-max(max(2*level-borderl[i], 2*level-borderr[i]), -1)-1);

	cout << ans << endl;

//	for(int i = begx + 1; i < endx; i++) cout << min(borderl[i], borderr[i]) << " " << rest[i] << endl;

}

int main() {

	//freopen("input.txt", "r", stdin);

	int t; cin >> t;

	while(t--) {

		init();

		solve();

	}

}