1198_E. Rectangle Painting 2
There is a square grid of size n × n. Some cells are colored in black, all others are colored in white. In one operation you can select some rectangle and color all its cells in white. It costs min(h, w) to color a rectangle of size h × w. You are to make all cells white for minimum total cost.

The square is large, so we give it to you in a compressed way. The set of black cells is the union of m rectangles.

Input

The first line contains two integers n and m (1 ≤ n ≤ 10^{9}, 0 ≤ m ≤ 50) — the size of the square grid and the number of black rectangles.

Each of the next m lines contains 4 integers x_{i1} y_{i1} x_{i2} y_{i2} (1 ≤ x_{i1} ≤ x_{i2} ≤ n, 1 ≤ y_{i1} ≤ y_{i2} ≤ n) — the coordinates of the bottom-left and the top-right corner cells of the i-th black rectangle.

The rectangles may intersect.

Output

Print a single integer — the minimum total cost of painting the whole square in white.

Examples

Input


10 2
4 1 5 10
1 4 10 5


Output


4


Input


7 6
2 1 2 1
4 2 4 3
2 5 2 5
2 3 5 3
1 2 1 2
3 2 5 3


Output


3

Note

The examples and some of optimal solutions are shown on the pictures below.

<image>
```python
import sys
from collections import defaultdict

class MaxFlow(object):
    def __init__(self):
        self.edges = defaultdict(lambda: defaultdict(lambda: 0))

    def add_edge(self, u, v, capacity=float('inf')):
        self.edges[u][v] = capacity

    def bfs(self, s, t):
        open_q = [s]

        visited = set()
        parent = dict()
        while open_q:
            close_q = []
            for node in open_q:
                for v, capacity in self.edges[node].items():
                    if v not in visited and capacity > 0:
                        close_q.append(v)
                        parent[v] = node
                        visited.add(v)
                        if v == t:
                            result = []
                            n2 = v
                            n1 = node
                            while n1 != s:
                                result.append((n1, n2))
                                n2 = n1
                                n1 = parent[n1]
                            result.append((n1, n2))
                            return result

            open_q = close_q

        return None

    def solve(self, s, t):
        flow = 0
        route = self.bfs(s, t)
        while route is not None:
            new_flow = float('inf')
            for _, (n1, n2) in enumerate(route):
                new_flow = min(new_flow, self.edges[n1][n2])
            for _, (n1, n2) in enumerate(route):
                self.edges[n1][n2] -= new_flow
                self.edges[n2][n1] += new_flow
            flow += new_flow

            route = self.bfs(s, t)

        return flow

    def __str__(self):
        result = "{ "
        for k, v in self.edges.items():
            result += str(k) + ":" + str(dict(v)) + ", "
        result += "}"
        return result


def main():
    (n, m) = tuple([int(x) for x in input().split()])
    r = []
    xs = set()
    ys = set()
    for i in range(m):
        (x1, y1, x2, y2) = tuple(int(x) for x in input().split())
        r.append((x1, y1, x2, y2))
        xs.add(x1)
        xs.add(x2 + 1)
        ys.add(y1)
        ys.add(y2 + 1)

    xx = sorted(xs)
    yy = sorted(ys)
    xsize = len(xs)
    ysize = len(ys)
    grid = []
    for i in range(ysize):
        grid.append([False] * xsize)

    for rect in r:
        x1 = rect[0]
        y1 = rect[1]
        x2 = rect[2]
        y2 = rect[3]
        for i, y in enumerate(yy):
            for j, x in enumerate(xx):
                if x1 <= x and y1 <= y and x2 >= x and y2 >= y:
                    grid[i][j] = True

    f = MaxFlow()
    for i in range(len(yy)):
        for j in range(len(xx)):
            if grid[i][j]:
                f.add_edge(1 + i, len(yy) + 1 + j, float('inf'))
    for i in range(len(yy) - 1):
        f.add_edge(0, i + 1, yy[i + 1] - yy[i])
    for i in range(len(xx) - 1):
        f.add_edge(len(yy) + 1 + i, len(xx) + len(yy) + 1, xx[i + 1] - xx[i])

    # print(xx)
    # print(yy)
    # print(f)
    print(f.solve(0, len(xx) + len(yy) + 1))


if __name__ == '__main__':
    main()

```