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++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
#include<string.h>
using namespace std;
#define FAST ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0)
#define endl "\n"
#define Max(x,y,z) max(x,max(y,z))
#define Min(x,y,z) min(x,min(y,z))
#define fr(i,s,e) for(i=s;i<e;i++)
#define rf(i,s,e) for(i=s-1;i>=e;i--)
#define pb push_back
#define eb emblace_back
#define mp make_pair
#define ff first
#define ss second
//#define int   long long
#define ll   long long
#define trace1(x)                cerr<<#x<<": "<<x<<endl
#define trace2(x, y)             cerr<<#x<<": "<<x<<" | "<<#y<<": "<<y<<endl
#define trace3(x, y, z)          cerr<<#x<<":" <<x<<" | "<<#y<<": "<<y<<" | "<<#z<<": "<<z<<endl
#define trace4(a, b, c, d)       cerr<<#a<<": "<<a<<" | "<<#b<<": "<<b<<" | "<<#c<<": "<<c<<" | "<<#d<<": "<<d<<endl
#define trace5(a, b, c, d, e)    cerr<<#a<<": "<<a<<" | "<<#b<<": "<<b<<" | "<<#c<<": "<<c<<" | "<<#d<<": "<<d<<" | "<<#e<< ": "<<e<<endl
#define trace6(a, b, c, d, e, f) cerr<<#a<<": "<<a<<" | "<<#b<<": "<<b<<" | "<<#c<<": "<<c<<" | "<<#d<<": "<<d<<" | "<<#e<< ": "<<e<<" | "<<#f<<": "<<f<<endl
#define vl vector<long long>
#define inf (1LL<<61)
#define vi vector<int>
#define vii vector< vector<int> >
#define vll vector< vector<long long> >
#define vpi vector< pair<int,int> >
#define vpl vector< pair<ll,ll> >
#define viii vector< vector<vector<int> > >
typedef pair<pair<int, int>,int> P;
#define cd complex<double>
#define PI 4*atan(1)
#define mod %1000000007
#define pi 3.14159265358979323846//20 decimals
#define ld long double
#define M 100000008

int main()
{
	FAST;
	ll n, m, i, j;
	cin >> n >> m;
	pair <ll ,ll> p[m];
	fr(i,0,m)
		cin >> p[i].ss >> p[i].ff;
	sort(p, p+m);
	ll t1 = p[0].ff;
	ll z = 0, mx = -1;
	fr(i,0,m)
		if(p[i].ss > mx)
			mx = p[i].ss;
	if(mx > t1)
		cout << "0";
	else
		cout << t1 - mx + 1;	
	
}