Problem:
Score: 600 points

Problem Statement

There is a grass field that stretches infinitely.

In this field, there is a negligibly small cow. Let (x, y) denote the point that is x\ \mathrm{cm} south and y\ \mathrm{cm} east of the point where the cow stands now. The cow itself is standing at (0, 0).

There are also N north-south lines and M east-west lines drawn on the field. The i-th north-south line is the segment connecting the points (A_i, C_i) and (B_i, C_i), and the j-th east-west line is the segment connecting the points (D_j, E_j) and (D_j, F_j).

What is the area of the region the cow can reach when it can move around as long as it does not cross the segments (including the endpoints)? If this area is infinite, print INF instead.

Constraints

All values in input are integers between -10^9 and 10^9 (inclusive).

1 \leq N, M \leq 1000

A_i < B_i\ (1 \leq i \leq N)

E_j < F_j\ (1 \leq j \leq M)

The point (0, 0) does not lie on any of the given segments.

Input

Input is given from Standard Input in the following format:

N M
A_1 B_1 C_1
:
A_N B_N C_N
D_1 E_1 F_1
:
D_M E_M F_M

Output

If the area of the region the cow can reach is infinite, print INF; otherwise, print an integer representing the area in \mathrm{cm^2}.

(Under the constraints, it can be proved that the area of the region is always an integer if it is not infinite.)

Sample Input 1

5 6
1 2 0
0 1 1
0 2 2
-3 4 -1
-2 6 3
1 0 1
0 1 2
2 0 2
-1 -4 5
3 -2 4
1 2 4

Sample Output 1

13

The area of the region the cow can reach is 13\ \mathrm{cm^2}.

Sample Input 2

6 1
-3 -1 -2
-3 -1 1
-2 -1 2
1 4 -2
1 4 -1
1 4 1
3 1 4

Sample Output 2

INF

The area of the region the cow can reach is infinite.

Solution in C++ (GCC 9.2.1) (judge verdict: Accepted)

//ほとんどbetrue12さんの提出(https://atcoder.jp/contests/abc168/submissions/13352386)の写経です

#pragma GCC optimize("Ofast")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx")
#pragma GCC optimize("unroll-loops")
//#pragma warning(disable : 4996)

#ifdef _MSC_VER
#include <intrin.h>
#define __builtin_popcount __popcnt
#define __builtin_popcountll __popcnt64
#endif

#include <limits.h>
#include <math.h>
#include <time.h>

#include <algorithm>
#include <array>
#include <bitset>
#include <cassert>
#include <complex>
#include <cstdio>
#include <cstring>
#include <functional>
#include <iomanip>
#include <iostream>
#include <iterator>
#include <map>
#include <numeric>
#include <queue>
#include <random>
#include <set>
#include <stack>
#include <string>
#include <unordered_map>
#include <unordered_set>
#include <vector>

using namespace std;

#define REP(i, n) for (int i = 0; i < (n); ++i)
#define REPR(i, n) for (int i = n - 1; i >= 0; --i)
#define FOR(i, m, n) for (int i = m; i < n; ++i)
#define FORR(i, m, n) for (int i = m - 1; i >= n; --i)
#define SORT(v, n) sort(v, v + n);
#define VSORT(v) sort(v.begin(), v.end());
#define REVERSE(v, n) reverse(v, v + n);
#define VREVERSE(v) reverse(v.begin(), v.end())
#define ll long long
#define print(x) cout << (x) << '\n'
#define pe(x) cout << (x) << " "
#define DEBUG(x) cout << #x << ": " << x << endl
#define lb(v, n) lower_bound(v.begin(), v.end(), (n))
#define ub(v, n) upper_bound(v.begin(), v.end(), (n))
#define int long long
#define double long double
#define all(x) (x).begin(), (x).end()
#define print_space(v) REP(i, v.size()) cout << v[i] << ((i == v.size() - 1) ? "\n" : " ")
template <typename T1, typename T2> inline void chmin(T1& a, T2 b) {
	if (a > b) a = b;
}
template <typename T1, typename T2> inline void chmax(T1& a, T2 b) {
	if (a < b) a = b;
}
typedef pair<int, int> pii;
typedef pair<long long, long long> pll;
typedef array<int, 3> arr3;
std::random_device rd;
std::mt19937 mt(rd());
constexpr ll MOD = 998244353;
constexpr int MAX = 500050;
const double pi = acos(-1);
constexpr double EPS = 1e-8;
constexpr ll LINF = 1e18 + 1;
constexpr int INF = 2e9;


template<typename T>
vector<T> compress(vector<T> A) {
	sort(A.begin(), A.end());
	A.erase(unique(A.begin(), A.end()), A.end());
	return A;
}

struct UnionFind {
	int N;
	vector<int> siz;
	vector<int> par;
	vector<int> Rank;
	UnionFind(int n) : siz(n, 1), par(n), Rank(n, 0) {
		N = n;
		for (int i = 0; i < N; i++) par[i] = i;
	}
	int find(int x) {
		if (par[x] == x) {
			return x;
		}
		else {
			return par[x] = find(par[x]);
		}
	}
	void unite(int x, int y) {
		x = find(x);
		y = find(y);
		if (x == y) return;
		if (Rank[x] < Rank[y]) {
			par[x] = y;
			siz[x] += siz[y];
			siz[y] = siz[x];
		}
		else {
			siz[y] += siz[x];
			siz[x] = siz[y];
			par[y] = x;
			if (Rank[x] == Rank[y]) Rank[x]++;
		}
	}
	bool same(int x, int y) { return find(x) == find(y); }
};

void solve() {
	int N, M; cin >> N >> M;
	vector<int>xs = { -INF,INF }, ys = { -INF,INF };
	vector<int>A(N), B(N), C(N), D(M), E(M), F(M);
	REP(i, N) {
		cin >> A[i] >> B[i] >> C[i];
		xs.push_back(A[i]);
		xs.push_back(B[i]);
		ys.push_back(C[i]);
	}
	REP(i, M) {
		cin >> D[i] >> E[i] >> F[i];
		xs.push_back(D[i]);
		ys.push_back(E[i]);
		ys.push_back(F[i]);
	}
	xs = compress<int>(xs);
	ys = compress<int>(ys);
	int xsz = xs.size(), ysz = ys.size();
	// リアル座標→圧縮座標
	auto find_x = [&](int a)->int {return ub(xs, a) - xs.begin() - 1; };
	auto find_y = [&](int a)->int {return ub(ys, a) - ys.begin() - 1; };
	// 圧縮後の座標の真上に線がある/真左に線がある
	static bitset<3100>above[3100], left[3100];
	// 線を引く
	REP(i, N) {
		A[i] = find_x(A[i]);
		B[i] = find_x(B[i]);
		C[i] = find_y(C[i]);
		FOR(x, A[i], B[i])left[x][C[i]] = 1;
	}
	REP(i, M) {
		D[i] = find_x(D[i]);
		E[i] = find_y(E[i]);
		F[i] = find_y(F[i]);
		FOR(y, E[i], F[i])above[D[i]][y] = 1;
	}
	//一次元に潰す
	auto encode = [&](int i, int j) {return i * ysz + j; };
	UnionFind uf(xsz*ysz);

	//線が引いていないところを連結
	FOR(i, 1, xsz)FOR(j,1, ysz) {
		if (!above[i][j])uf.unite(encode(i, j), encode(i - 1, j));
		if (!left[i][j])uf.unite(encode(i, j), encode(i, j - 1));
	}

	// 牛と同じ連結成分について面積を足す 無限遠と連結ならINF
	int cow = encode(find_x(0), find_y(0));
	int ans = 0;
	auto fail = []() {
		print("INF");
		exit(0);
	};
	REP(i, xsz)REP(j, ysz)if (uf.same(cow, encode(i, j))) {
		if (i == 0 || i == xsz - 1 || j == 0 || j == ysz - 1)fail();
		int dx = xs[i + 1] - xs[i], dy = ys[j + 1] - ys[j];
		ans += dx * dy;
	}
	print(ans);
}




signed main() {
	cin.tie(0);
	ios::sync_with_stdio(false);
	//mt.seed((int)(time(0)));
	//REP(_, 10000)test();
	//int q;
	//cin >> q;
	//while (q--)
	solve();
}