//習題5-6　對稱軸（Symmetry, ACM/ICPC Seoul 2004, UVa1595）
/*
定義 map<pair<int, int>, int> vis
插入時賦值為1 vis[pair(x, y)]++;
查詢一個關鍵字是否存在：if(!vis[pair<int, int>(x, y)]) ... 如果不存在會新建立一個關鍵字，返回值為0
是否可以等同於if(!vis.count(pair<int, int>(x, y))) ...呢？
*/
struct A {
	int x, y;
	bool operator<(const A &rhs) {
		return x < rhs.x;
	}
}a[1005];
using PII = pair<int, int>;
PII p[1005];
map<PII, int> vis;

int main() {
	int T, n;
	//IN(); OUT();
	scanf("%d", &T);
	while (T--) {
		vis.clear();
		scanf("%d", &n);
		for (int i = 0; i < n; ++i) {
			scanf("%d%d", &a[i].x, &a[i].y);
			p[i] = make_pair(a[i].x, a[i].y);
			if (!vis.count(p[i])) vis[p[i]]++;
		}
		sort(a, a + n);
		int mid = n / 2, s;
		if (n % 2) s = 2 * a[mid].x;
		else s = (a[mid].x + a[mid - 1].x);
		//printf("s = %d\n", s);
		int x, y;
		bool flag = true;
		for (int i = 0; i < n; i++) {
			x = s - a[i].x;
			y = a[i].y;
			//printf("(%d, %d) -> (%d, %d)\n",a[i].x, a[i].y, x, y);
			if (!vis.count(PII(x, y))) {
				flag = false;
				break;
			}
		}
		if (flag) printf("YES\n");
		else printf("NO\n");
	}
	return 0;
}