# 浮点数的符号函数
ESP = 10 ** (-8)
def sign(val: float):
    if abs(val) < ESP:
        return 0
    return 1 if val > 0 else -1


# 向量的外乘积，表示两个向量围成的平行四边形面积，b在a逆时针方向，面积为正，否则面积为负
# 输入是两个向量的坐标(x1, y1), (x2, y2)
def cross(a, b):
    return a[0]*b[1] - b[0]*a[1]

# 判断两线段是否相交, a1, a2是线段1两个端点，b1, b2是线段2两个端点
def has_segment_intersection(a1, a2, b1, b2)->bool:
    c1 = cross( (a2[0]-a1[0], a2[1]-a1[1]), (b1[0]-a1[0], b1[1]-a1[1]) )
    c2 = cross( (a2[0]-a1[0], a2[1]-a1[1]), (b2[0]-a1[0], b2[1]-a1[1]) )
    c3 = cross( (b2[0]-b1[0], b2[1]-b1[1]), (a2[0]-b1[0], a2[1]-b1[1]) )
    c4 = cross( (b2[0]-b1[0], b2[1]-b1[1]), (a1[0]-b1[0], a1[1]-b1[1]) )
    return sign(c1) * sign(c2) <= 0 and sign(c3) * sign(c4) <= 0


# 求解两条直线的交点(注意前提是两条直线不平行)
# p1, v1 是第一条直线的点向式表示，p2, v2 是第二条直线的点向式表示
def intersection_point(p1, v1, p2, v2):
    u = (p1[0]-p2[0], p1[1]-p2[1])

    if sign(cross(v1, v2)) == 0:
        return None

    t = cross(v2, u) / cross(v1, v2)
    return (p1[0] + v1[0]*t, p1[1] + v1[1]*t)


# 求解点到线段的交点, a1, b1 是第一条线段的两个端点，a2, b2是第二条线段两个端点
def intersection_point_on_segment(a1, b1, a2, b2):
    if not has_segment_intersection(a1, b1, a2, b2):
        return None

    v1 = (b1[0]-a1[0], b1[1]-a1[1])
    v2 = (b2[0]-a2[0], b2[1]-a2[1])
    return intersection_point(a1, v1, a2, v2)


n = int(input())
seg = []
triangle = []

x_set = set()
for _ in range(n):
    x1, y1, x2, y2, x3, y3 = map(float, input().split())
    triangle.append((x1, y1, x2, y2, x3, y3))
    a, b, c = (x1, y1), (x2, y2), (x3, y3)
    seg.append((a, b))
    seg.append((a, c))
    seg.append((b, c))
    x_set.add(x1)
    x_set.add(x2)
    x_set.add(x3)

# 求所有线段的交点
for i in range(len(seg)):
    for j in range(i+1, len(seg)):
        pp = intersection_point_on_segment(seg[i][0], seg[i][1], seg[j][0], seg[j][1])
        if pp is None:
            continue
        x_set.add(pp[0])

x_pos = list(x_set)
x_pos.sort()

# x坐标是pos位置的扫描线做区间合并
from functools import lru_cache
@lru_cache(typed=False, maxsize=128000000)
def merge_range(pos):
    rr = []
    for x1, y1, x2, y2, x3, y3 in triangle:
        if pos < min(x1, x2, x3) or pos > max(x1, x2, x3):
            continue

        pp1 = intersection_point_on_segment((x1, y1), (x2, y2), (pos, -10000000), (pos, 10000000))
        pp2 = intersection_point_on_segment((x1, y1), (x3, y3), (pos, -10000000), (pos, 10000000))
        pp3 = intersection_point_on_segment((x3, y3), (x2, y2), (pos, -10000000), (pos, 10000000))
        y_arr = []
        for pp in [pp1, pp2, pp3]:
            if pp is not None:
                y_arr.append(pp[1])

        if len(y_arr) < 2:
            continue

        y_arr.sort()
        if len(y_arr) == 2:
            rr.append((y_arr[0], y_arr[1]))
        else:
            rr.append((y_arr[0], y_arr[1]))
            rr.append((y_arr[1], y_arr[2]))

    # 区间合并
    rr.sort()
    tot = 0
    max_end = -0x7fffffff  # 当前已经枚举过的区间的结尾的最大值
    for a, b in rr:
        if a <= max_end:
            if b > max_end:
                tot += b - max_end
                max_end = b
        else:
            tot += b - a
            max_end = b

    return tot

# 枚举区间
ans = 0
for ii in range(len(x_pos)-1):
    left, right = x_pos[ii], x_pos[ii+1]

    sum1, sum2 = merge_range(left), merge_range(right)
    # print(left, right, sum1, sum2)

    ans += (sum1 + sum2) * (right - left) / 2

print("%.2f" % ans)