Problem:
Score : 300 points

Problem Statement

We have N ID cards, and there are M gates.

We can pass the i-th gate if we have one of the following ID cards: the L_i-th, (L_i+1)-th, ..., and R_i-th ID cards.

How many of the ID cards allow us to pass all the gates alone?

Constraints

All values in input are integers.

1 \leq N \leq 10^5

1 \leq M \leq 10^5

1 \leq L_i \leq R_i \leq N

Input

Input is given from Standard Input in the following format:

N M
L_1 R_1
L_2 R_2
\vdots
L_M R_M

Output

Print the number of ID cards that allow us to pass all the gates alone.

Sample Input 1

4 2
1 3
2 4

Sample Output 1

2

Two ID cards allow us to pass all the gates alone, as follows:

The first ID card does not allow us to pass the second gate.

The second ID card allows us to pass all the gates.

The third ID card allows us to pass all the gates.

The fourth ID card does not allow us to pass the first gate.

Sample Input 2

10 3
3 6
5 7
6 9

Sample Output 2

1

Sample Input 3

100000 1
1 100000

Sample Output 3

100000

Solution in Java8 (OpenJDK 1.8.0) (judge verdict: Wrong Answer)

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.AbstractMap;
import java.util.Arrays;
import java.util.StringTokenizer;

public class Main {

	public static void main(String[] args) {
		FastScanner sc = new FastScanner(System.in);
		PrintWriter out = new PrintWriter(System.out);

		solve(sc,out);

		out.flush();
	}

	public static void solve(FastScanner sc, PrintWriter out) {
		int n = sc.nextInt();
		int m = sc.nextInt();

		int[] l = new int[m];
		int[] r = new int[m];

		for(int i = 0; i < m; i++){
			l[i] = sc.nextInt();
			r[i] = sc.nextInt();
		}
		
		int[] x = new int[n];
		
		for(int i = 0; i < m; i++) {
			x[l[i]-1]++;
			x[r[i]-1]--;
		}
		
		int res = 0;
		
		for(int i = 0; i < n; i++) {
			if(0 < i) x[i] += x[i -1];
		}
		
		Arrays.sort(x);
		
		int max = x[x.length-1];
		
		for(int i = x.length-1; i >= 0; i--) {
			if(x[i] == max) {
				res++;
			}else{
				break;
			}
		}
		
		
		out.println(res+1);
		
 	}

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

class Pair<K, V> extends AbstractMap.SimpleEntry<K, V> {

    public Pair(final K key, final V value) {
        super(key, value);
    }
 }

class FastScanner {
    private BufferedReader reader = null;
    private StringTokenizer tokenizer = null;

    public FastScanner(InputStream in) {
        reader = new BufferedReader(new InputStreamReader(in));
        tokenizer = null;
    }

    public String next() {
        if (tokenizer == null || !tokenizer.hasMoreTokens()) {
            try {
                tokenizer = new StringTokenizer(reader.readLine());
            } catch (IOException e) {
                throw new RuntimeException(e);
            }
        }
        return tokenizer.nextToken();
    }

    public String nextLine() {
        if (tokenizer == null || !tokenizer.hasMoreTokens()) {
            try {
                return reader.readLine();
            } catch (IOException e) {
                throw new RuntimeException(e);
            }
        }

        return tokenizer.nextToken("\n");
    }

    public long nextLong() {
        return Long.parseLong(next());
    }

    public int nextInt() {
        return Integer.parseInt(next());
    }

    public double nextDouble() {
         return Double.parseDouble(next());
     }

    public int[] nextIntArray(int n) {
        int[] a = new int[n];
        for (int i = 0; i < n; i++)
            a[i] = nextInt();
        return a;
    }

    public long[] nextLongArray(int n) {
        long[] a = new long[n];
        for (int i = 0; i < n; i++)
            a[i] = nextLong();
        return a;
    }
}