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 C# (Mono 4.6.2.0) (judge verdict: Time Limit Exceeded)

using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
using System.Threading.Tasks;
using System.Text.RegularExpressions;
using static System.Console;

namespace AtCoder_1
{
    class Program
    {
        static void Main(string[] args)
        {
            var t = CinTAr();
            int[] f = new int[t[0]];
            for (int i = 0; i < t[1]; i++)
            {
                var g = CinTAr();
                for (int af = g[0] - 1; af <= g[1] -1; af++)
                {
                        f[af]++;
                }
            }
            int count = 0;
            for (int i = 0; i < f.Length; i++)
            {
                if (f[i] == t[1])
                {
                    count++;
                }
            }
            Cout(count);
            Cin();
        }



        static readonly int MOD1E9 = (int)1e9 + 7;
        static string Cin() { return ReadLine(); }
        static string[] CinAr() { return ReadLine().Split(' '); }
        static int CinT() { return S2i(Cin()); }
        static int[] CinTAr() { return CinAr().Select((x, i) => { return S2i(x); }).ToArray(); }
        static long CinL() { return S2L(Cin()); }
        static long[] CinLAr() { return CinAr().Select((x, i) => { return S2L(x); }).ToArray(); }
        static double CinD() { return S2D(Cin()); }
        static double[] CinDAr() { return CinAr().Select((x, i) => { return S2D(x); }).ToArray(); }
        static int S2i(string s) { return int.Parse(s); }
        static long S2L(string s) { return long.Parse(s); }
        static double S2D(string s) { return double.Parse(s); }
        static void OYes() { Cout("Yes"); }
        static void ONo() { Cout("No"); }
        static void Cout(object s) { WriteLine(s); }
        static int[] yakusuu(int n)
        {
            if (n == 1) { return new int[] { 1 }; }
            HashSet<int> res = new HashSet<int>();
            for (int i = 1; i < Math.Ceiling(n / 2.0); i++)
            {
                if (n % i == 0)
                {
                    res.Add(i);
                    res.Add(n / i);
                }
            }
            return res.ToArray();
        }
        /// <summary>
        /// Debug時のみの使用を推奨
        /// </summary>
        /// <typeparam name="Type"></typeparam>
        /// <param name="types"></param>
        static void CoutAr<Type>(IEnumerable<Type> types)
        {
            string text = "";
            foreach (var i in types)
            {
                text += " " + i;
            }
            WriteLine(text.Substring(1));
        }

        public static long nCr(int n, int r)
        {
            // naive: return Factorial(n) / (Factorial(r) * Factorial(n - r));
            return nPr(n, r) / Factorial(r);
        }

        public static long nPr(int n, int r)
        {
            // naive: return Factorial(n) / Factorial(n - r);
            return FactorialDivision(n, n - r);
        }

        private static long FactorialDivision(int topFactorial, int divisorFactorial)
        {
            long result = 1;
            for (int i = topFactorial; i > divisorFactorial; i--)
                result *= i;
            return result;
        }

        private static long Factorial(int i)
        {
            if (i <= 1)
                return 1;
            return i * Factorial(i - 1);
        }
    }
}