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)

#pragma gcc optimize("Ofast")
#include <bits/stdc++.h>

using namespace std;
using ll = long long;

#define FOR(i, a, b) for (int i = (a); i < (b); ++i)
#define REP(i, n) FOR(i, 0, n)
#define ALL(V) (V).begin(),(V).end()
#define SORT(V) sort(ALL(V)) //小さい方からソート
#define REV(V) reverse(ALL(V)) //リバース
#define RSORT(V) SORT(V);REV(V) //大きい方からソート
#define NPN(V) next_permutation(ALL(V)) //順列
#define pb(n) push_back(n)
#define endl '\n'
#define Endl '\n'
#define DUMP(x)  cout << #x << " = " << (x) << endl
#define YES(n) cout << ((n) ? "YES" : "NO"  ) << endl
#define Yes(n) cout << ((n) ? "Yes" : "No"  ) << endl
#define yes(n) cout << ((n) ? "yes" : "no"  ) << endl
#define Yay(n) cout << ((n) ? "Yay!": ":(") << endl
#define RAPID cin.tie(0);ios::sync_with_stdio(false)
#define VSUM(V) accumulate(ALL(V), 0)
#define MID(a,b,c) (a) < (b) && (b) < (c)
#define MIDe(a,b,c) (a) <= (b) && (b) <= (c)
#define IN(n) cin >> n
#define IN2(a,b) cin >> a >> b
#define IN3(a,b,c) cin >> a >> b >> c 
#define IN4(a,b,c,d) cin >> a >> b >> c >> d
#define VIN(V) for(int i = 0; i < (V).size(); i++) {cin >> (V).at(i);}
#define OUT(n) cout << n << endl
#define VOUT(V) REP(i, (V).size()) {cout << (V)[i] << endl;}
#define VOUT2(V) REP(i, (V).size()) {cout << (V)[i] << " ";} cout<<endl;

// 型マクロ定義
#define int long long
#define P pair<ll,ll>
#define Vi vector<ll>
#define VVi vector<vector<ll>>
#define Vd vector<double>
#define Vb vector<bool>
#define Vs vector<string>
#define Vc vector<char>
#define M map<ll,ll>
#define S set<ll>
#define PQ priority_queue<ll>
#define PQG priority_queue<ll,V,greater<ll>
//

template<class T> inline bool chmin(T& a, T b) {if(a>b){a=b;return true;}return false;}
template<class T> inline bool chmax(T& a, T b) {if(a<b){a=b;return true;}return false;}

const int MOD = 1000000007;
const int INF = 1061109567;
const double EPS = 1e-10;
const double PI  = acos(-1.0);

int gcd(int a,int b){return b != 0 ? gcd(b, a % b) : a;} //最大公約数
int lcm(int a, int b){return a * b / gcd(a, b);}
string toStrUp(string str){char diff = 'A'-'a';REP(i,str.size()) str[i] += diff;return str;}
string gCh(string str, int key){return str.substr(key,1);}

signed main() {
  RAPID;
  // デフォルト変数定義
  int n=0,m=0,a=0,b=0,c=0,d=0,x=0,y=0,z=0;
  string s="",t="";
  //

  // ここから
  IN2(n,m);
  Vi L(m);
  Vi R(m);
  REP(i,m){
    IN2(L[i],R[i]);
  }
  if(*min_element(ALL(R)) >= *max_element(ALL(L))) OUT(*min_element(ALL(R)) - *max_element(ALL(L)) + 1);
  else OUT(0);
}