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: Accepted)

import java.util.Scanner;

public class Main {

	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		int n = sc.nextInt();
		int m = sc.nextInt();
		int max = 1;//lの最大値格納先
		int min = n;//Rの最小値格納先
		int l;
		int r;
		for (int i = 1; i <= m; i++) {
			l = sc.nextInt();
			r = sc.nextInt();//ｌとｒを取得

			if (l > max) {
				max = l;
			}
			if (r < min) {
				min = r;
			}//普通にlとrの最大、最小をいつものやり方で出す
		}
		//lとrの最大、最小の差を取り、1を足す(2～6の鍵がOKの時、鍵は5本だが6-2=4になってしまうのでその分の調整)

		int ans = min - max + 1;
		if (ans >= 0) {//ansが正ならそのまま表示
			System.out.println(ans);
		} else {//ansが負なら鍵は存在しないので0を表示
			System.out.println(0);
		}
	}
}