Problem: 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>

Solution (in JAVA):

Solution:
import java.io.BufferedInputStream;
import java.io.BufferedOutputStream;
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.io.StreamTokenizer;
import java.util.Arrays;
import java.util.LinkedList;
import java.util.Queue;
import java.util.Scanner;

public class Main {
	static StreamTokenizer st = new StreamTokenizer(new BufferedInputStream(System.in));
	static BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
	static PrintWriter pr = new PrintWriter(new BufferedOutputStream(System.out));
	static Scanner sc = new Scanner(System.in);

//		long tic = System.currentTimeMilis();
//		long toc = System.currentTimeMillis();
//		System.out.println("Elapsed time: " + (toc - tic) + " ms");
	/**
	 * a[]、b[] 分别代表离散化的行和列 c[]代表离散简化后的图。
	 * x1[]、x2[]、y1[]、y2[] ,代表对应的黑色方块。
	 * ss源点 tt汇点
	 * e[]代表二分图中的边,s Node.f代表容量
	 */
	static Node e[] = new Node[10000001];
	static int et = 1, h[] = new int[10001], d[] = new int[10001], ss, tt;
	static int a[] = new int[101], b[] = new int[101], c[][] = new int[101][101];
	static int x1[] = new int[101], y1[] = new int[101], x2[] = new int[101], y2[] = new int[101];

	public static void main(String[] args) {
		int n = nextInt(), m = nextInt();
		for (int i = 1; i <= m; i++) {
			x1[i] = nextInt();
			y1[i] = nextInt();
			x2[i] = nextInt();
			y2[i] = nextInt();
			a[++a[0]] = x1[i];
			a[++a[0]] = x2[i] + 1;
			b[++b[0]] = y1[i];
			b[++b[0]] = y2[i] + 1;
		}
		Arrays.sort(a, 1, a[0] + 1);
		a[0] = unique(a, 1, a[0] + 1);
 
		Arrays.sort(b, 1, b[0] + 1);
		b[0] = unique(b, 1, b[0] + 1) ;
		for (int i = 1; i <= m; i++) {
			x1[i] = lower_bound(a, 1, a[0] + 1, x1[i]);
			x2[i] = lower_bound(a, 1, a[0] + 1, x2[i] + 1) - 1;
			y1[i] = lower_bound(b, 1, b[0] + 1, y1[i]);
			y2[i] = lower_bound(b, 1, b[0] + 1, y2[i] + 1) - 1;
			for (int u = x1[i]; u <= x2[i]; u++)
				for (int v = y1[i]; v <= y2[i]; v++)
					c[u][v] = 1;
		}
		ss = 0;
		tt = a[0] + b[0] + 1;
		for (int i = 1; i <= a[0]; i++)
			for (int j = 1; j <= b[0]; j++)
				if (c[i][j] != 0)
					jb(i, a[0] + j, (int) 2e9);
		for (int i = 1; i < a[0]; i++)
			jb(ss, i, a[i + 1] - a[i]);
		for (int i = 1; i < b[0]; i++)
			jb(a[0] + i, tt, b[i + 1] - b[i]);
		int ans = 0;
		while (bfs()) {
			ans += dfs(ss, (int) 2e9);
		}
		System.out.printf("%d\n", ans);
	}
 
	private static int lower_bound(int[] arr, int l, int r, int it) {
		int m = (l+r)>>1;
		while(l<=r) {
			if(it > arr[m]) {
				l = m+1;
				m = (l+r)>>1;
			}else if(it<arr[m]) {
				r = m-1;
				m = (l+r)>>1;
			}else {
				return m;
			}
		}
		return m;
	}
 
	private static int unique(int[] a2, int a, int b) {
		int te = a;
		for (int i = a + 1; i < b; i++) {
			if (a2[i] != a2[te]) {
				a2[++te] = a2[i];
			}
		}
		return te - a + 1;
	}
 
	static void jb(int u, int v, int f) {
		e[++et] = new Node(v, f, h[u]);
		h[u] = et;
		e[++et] = new Node(u, 0, h[v]);
		h[v] = et;
	}
 
	static boolean bfs() {
		for (int i = ss; i <= tt; i++) {
			d[i] = 0;
		}
		Queue<Integer> q = new LinkedList<Integer>();
		q.add(ss);
		d[ss] = 1;
		while (!q.isEmpty()) {
			int u = q.poll();
			for (int i = h[u]; i != 0; i = e[i].nx)
				if (e[i].f != 0 && d[e[i].v] == 0) {
					d[e[i].v] = d[u] + 1;
					q.add(e[i].v);
				}
		}
		return d[tt] != 0;
	}
 
	static int dfs(int u, int f) {
		if (f == 0 || u == tt)
			return f;
		int F = 0;
		for (int i = h[u]; i != 0; i = e[i].nx)
			if (d[e[i].v] == d[u] + 1) {
				int ff = dfs(e[i].v, Math.min(e[i].f, f));
				e[i].f -= ff;
				e[i ^ 1].f += ff;
				f -= ff;
				F += ff;
				if (f == 0)
					break;
			}
		return F;
	}
 
	static long pow(long a, long b, long mod) {
		if (b == 0)
			return 1;
		if (b == 1)
			return a % mod;
		if ((b & 1) == 0)
			return pow(a * a % mod, b >> 1, mod) % mod;
		else
			return a * pow(a * a % mod, b >> 1, mod) % mod;
	}

	static long gcd(long a, long b) {
		if (b == 0) {
			return a;
		} else {
			return gcd(b, a % b);
		}
	}

	static int nextInt() {
		try {
			st.nextToken();
		} catch (IOException e) {
			e.printStackTrace();
		}
		return (int) st.nval;
	}

	static double nextDouble() {
		try {
			st.nextToken();
		} catch (IOException e) {
			e.printStackTrace();
		}
		return st.nval;
	}

	static String next() {
		try {
			st.nextToken();
		} catch (IOException e) {
			e.printStackTrace();
		}
		return st.sval;
	}

	static long nextLong() {
		try {
			st.nextToken();
		} catch (IOException e) {
			e.printStackTrace();
		}
		return (long) st.nval;
	}
}

class Node {
	int v, f, nx;

	public Node() {
	}

	public Node(int v, int f, int nx) {
		this.v = v;
		this.f = f;
		this.nx = nx;
	}

}