# Original solution
def f_single_dot():
    from collections import deque
    N, M = [int(i) for i in input().split()]

    x = {0}  # 重複を除いた x 座標の一覧 (座標圧縮に必要)
    y = {0}
    vertical_lines = []
    horizontal_lines = []
    for _ in range(N):
        a, b, c = [int(i) for i in input().split()]
        vertical_lines.append((a, b, c))
        x |= {a, b}
        y |= {c}
    for _ in range(M):
        d, e, f = [int(i) for i in input().split()]
        horizontal_lines.append((d, e, f))
        x |= {d}
        y |= {e, f}
    x = sorted(x)
    y = sorted(y)

    # 座標圧縮 (壁も配列に格納するので、2 倍している)
    x_comp = {x: 2 * k for k, x in enumerate(x)}
    y_comp = {y: 2 * k for k, y in enumerate(y)}
    # 座標圧縮されたグリッドで縦横に 1 マス分進むことは実際にはどれくらい進んだことなのか
    length_x = [x[i // 2 + 1] - x[i // 2] if i % 2 else 0 for i in range(len(x) * 2 - 1)]
    length_y = [y[i // 2 + 1] - y[i // 2] if i % 2 else 0 for i in range(len(y) * 2 - 1)]
    # print(x,y)
    # print(length_x,length_y)

    x_size, y_size = 2 * len(x) - 1, 2 * len(y) - 1
    grid = [[0] * y_size for _ in range(x_size)]

    # 壁を埋めていく (累積和を用いていない)
    for a, b, c in vertical_lines:
        a, b, c = x_comp[a], x_comp[b], y_comp[c]
        for x in range(a, b + 1):
            grid[x][c] = 1
    for d, e, f in horizontal_lines:
        d, e, f = x_comp[d], y_comp[e], y_comp[f]
        for y in range(e, f + 1):
            grid[d][y] = 1

    def is_inside(x, y):
        return (0 <= x < x_size) and (0 <= y < y_size)

    ans = 0
    diff = ((-1, 0), (1, 0), (0, -1), (0, 1))
    start_x, start_y = x_comp[0], y_comp[0]
    is_visited = [[False] * y_size for _ in range(x_size)]
    queue = deque([(start_x, start_y)])
    while queue:
        x, y = queue.pop()
        is_visited[x][y] = True
        if (x % 2) & (y % 2):
            ans += length_x[x] * length_y[y]
        for dx, dy in diff:
            nx, ny = x + dx, y + dy
            if not is_inside(nx, ny):
                return 'INF'  # グリッドの外へ出た ⇔ 囲いになっていない
            if (not is_visited[nx][ny]) and (grid[nx][ny] == 0):
                queue.appendleft((nx, ny))
                is_visited[nx][ny] = True
    return ans

print((f_single_dot()))

# Optimized solution
def f_single_dot():
    # https://atcoder.jp/contests/abc168/submissions/13368489
    import sys
    from collections import deque
    input = sys.stdin.readline
    N, M = [int(i) for i in input().split()]

    X, Y = set([0]), set([0])
    A, B, C = [0] * N, [0] * N, [0] * N
    for i in range(N):
        A[i], B[i], C[i] = list(map(int, input().split()))
        X.add(A[i])
        X.add(B[i])
        Y.add(C[i])
    D, E, F = [0] * M, [0] * M, [0] * M
    for i in range(M):
        D[i], E[i], F[i] = list(map(int, input().split()))
        X.add(D[i])
        Y.add(E[i])
        Y.add(F[i])
    UX = list(X)
    UY = list(Y)
    UX.sort()
    UY.sort()
    dictx, dicty = {n: i * 2 for i, n in enumerate(UX)}, {n: i * 2 for i, n in enumerate(UY)}
    LX = [UX[i // 2 + 1] - UX[i // 2] if i % 2 else 0 for i in range(len(UX) * 2 - 1)]
    LY = [UY[i // 2 + 1] - UY[i // 2] if i % 2 else 0 for i in range(len(UY) * 2 - 1)]
    c = [[0] * (len(UX) * 2 - 1) for _ in range(len(UY) * 2 - 1)]
    for i in range(N):
        y = dicty[C[i]]
        for x in range(dictx[A[i]], dictx[B[i]] + 1):
            c[y][x] = 1
    for i in range(M):
        x = dictx[D[i]]
        for y in range(dicty[E[i]], dicty[F[i]] + 1):
            c[y][x] = 1
    que = deque()
    used = [[0] * len(UX) * 2 for _ in range(len(UY) * 2)]
    dy = [1, 1, -1, -1]
    dx = [1, -1, 1, -1]
    for k in range(4):
        nny, nnx = dicty[0] + dy[k], dictx[0] + dx[k]
        if 1 <= nny < len(UY) * 2 - 1 and 1 <= nnx < len(UX) * 2 - 1:
            que.append((nny, nnx))
            used[nny][nnx] = 1
    ans = 0
    dy = [1, 0, -1, 0]
    dx = [0, 1, 0, -1]
    while que:
        y, x = que.popleft()
        if y % 2 and x % 2:
            ans += LY[y] * LX[x]
        for k in range(4):
            ny, nx = y + dy[k], x + dx[k]
            nny, nnx = ny + dy[k], nx + dx[k]
            if c[ny][nx] == 0:
                if nny < 0 or nny >= len(UY) * 2 - 1 or nnx < 0 or nnx >= len(UX) * 2 - 1:
                    return 'INF'
                if not used[nny][nnx]:
                    que.append((nny, nnx))
                    used[nny][nnx] = 1
    return ans

print((f_single_dot()))