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: Wrong Answer)

#define _USE_MATH_DEFINES
#include<bits/stdc++.h>
#define int long long
#define rep(i,n) for(int i=0;i<n;i++)
#define Rep(i,a,b) for(int i=0;i<b;i++)
#define REP(i,a,b) for(int i=0;i<=b;i++)
#define rev(i,n) for(int i=n-1;i>=0;i--)
#define vi vector<int>
#define vvi vector<vi>
#define pb push_back
#define pi pair<int,int>
#define vp vector<pair<int,int>>
#define mp make_pair
#define all(v) (v).begin(),(v).end()
#define fi first
#define se second
#define MEMSET(a) memset(a,0,sizeof(a))
#define inf (1ll<<60)
#define Yes(f) cout <<(f?"Yes":"No")<<endl//fが真の場合Yesを出力、偽の場合Noを出力
#define yes(f) cout <<(f?"yes":"no")<<endl
#define YES(f) cout <<(f?"YES":"NO")<<endl
#define SORT(v) sort(all(v))
#define RSORT(v) sort(all(v),greater<int>())

using namespace std;

const int mod=1e9+7;

signed main(){
    int n,m;
    cin >> n >> m;
    int fast=0,last=100000;
    int l,r;
    rep(i,m)
    {
        cin >> l >> r;
        fast=max(fast,l);
        last=min(last,r);
    }
    int ans;
    if(fast>last)ans=1;
    else ans=last-fast+1;
    cout << ans << endl;
}