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 Rust (1.15.1) (judge verdict: Wrong Answer)

#![allow(unused_imports)]
#![allow(non_snake_case)]
#![allow(dead_code)]
use std::cell::RefCell;
use std::cmp::{max, min, Ordering};
use std::collections::*;
use std::fmt::{Debug, Formatter, Write as FmtWrite};
use std::io::{stderr, stdin, BufRead, Write};
use std::mem::{replace, swap};
use std::ops::*;
use std::rc::Rc;
use std::usize;

const MOD_10_9_7: u64 = 1_000_000_007;

/// FYI: https://github.com/vain0x/scan-bench
#[allow(unused_macros)]
macro_rules! read {
    ([$t:ty] ; $n:expr) =>
        ((0..$n).map(|_| read!([$t])).collect::<Vec<_>>());
    ($($t:ty),+ ; $n:expr) =>
        ((0..$n).map(|_| read!($($t),+)).collect::<Vec<_>>());
    ([$t:ty]) =>
        (rl().split_whitespace().map(|w| w.parse().unwrap()).collect::<Vec<$t>>());
    ($t:ty) =>
        (rl().parse::<$t>().unwrap());
    ($($t:ty),*) => {{
        let buf = rl();
        let mut w = buf.split_whitespace();
        ($(w.next().unwrap().parse::<$t>().unwrap()),*)
    }};
}

fn rl() -> String {
    let mut buf = String::new();
    stdin().read_line(&mut buf).unwrap();
    buf.trim_right().to_owned()
}

////////////////////////////////////////////////////////////////////////////////////////////////////
/// 飛ばし飛ばしで値を使うイテレータ Rust v1.28以降でしか使えないため独自実装
/// 簡易実装のためインデックスのオーバーフロー等は考慮していない
pub struct StepBy<I> {
    iter: I,
    step: usize,
    first_take: bool,
}
impl<I> StepBy<I> {
    pub fn new(iter: I, step: usize) -> StepBy<I> {
        StepBy {
            iter: iter,
            step: step - 1,
            first_take: true,
        }
    }
}
impl<I> Iterator for StepBy<I>
where
    I: Iterator,
{
    type Item = I::Item;

    fn next(&mut self) -> Option<Self::Item> {
        if self.first_take {
            self.first_take = false;
            self.iter.next()
        } else {
            self.iter.nth(self.step)
        }
    }

    fn size_hint(&self) -> (usize, Option<usize>) {
        let inner_hint = self.iter.size_hint();
        if self.first_take {
            let f = |n| {
                if n == 0 {
                    0
                } else {
                    1 + (n - 1) / (self.step + 1)
                }
            };
            (f(inner_hint.0), inner_hint.1.map(f))
        } else {
            let f = |n| n / (self.step + 1);
            (f(inner_hint.0), inner_hint.1.map(f))
        }
    }
}
impl<I> ExactSizeIterator for StepBy<I> where I: ExactSizeIterator {}
////////////////////////////////////////////////////////////////////////////////////////////////////

trait IteratorExt: Iterator + Sized {
    // ref: https://qiita.com/vain0x/items/512784ff60ce599dccae#vec
    fn vec(self) -> Vec<Self::Item> {
        self.collect()
    }

    // step_by は v1.28.0 以降でしか使えないため独自実装
    fn stepby(self, step: usize) -> StepBy<Self>
    where
        Self: Sized,
    {
        StepBy::new(self, step)
    }
}

impl<T: Iterator> IteratorExt for T {}

////////////////////////////////////////////////////////////////////////////////////////////////////
/// 逆順ソート ReverseがRust v1.19以降でしか使えないため独自実装
#[derive(Eq, PartialEq, Clone, Debug)]
pub struct Rev<T>(pub T);

impl<T: PartialOrd> PartialOrd for Rev<T> {
    fn partial_cmp(&self, other: &Rev<T>) -> Option<Ordering> {
        other.0.partial_cmp(&self.0)
    }
}

impl<T: Ord> Ord for Rev<T> {
    fn cmp(&self, other: &Rev<T>) -> Ordering {
        other.0.cmp(&self.0)
    }
}
// 逆順ソートここまで
////////////////////////////////////////////////////////////////////////////////////////////////////

////////////////////////////////////////////////////////////////////////////////////////////////////
/// べき乗のmodを効率的に求める関数
/// ref: https://algorithmbeginner.blogspot.com/2018/02/blog-post_23.html
fn pow(x: u64, n: u64, modulo: u64) -> u64 {
    let mut res = 1;
    if n > 0 {
        res = pow(x, n / 2, modulo);
        if n % 2 == 0 {
            res = (res * res) % modulo;
        } else {
            res = (((res * res) % modulo) * x) % modulo;
        }
    }
    res
}
////////////////////////////////////////////////////////////////////////////////////////////////////

fn main() {
    let (n, m) = read!(usize, usize);
    let v: Vec<(usize, usize)> = read!(usize, usize; m);

    #[derive(Clone, Copy, Debug)]
    struct Range {
        start: usize,
        end: usize,
    };

    // vの要素のうち1つを e とすると [e.0, e.1] の区間内の整数であればゲートを通過できる
    // [e.0, e.1] の区間の共通集合を求め、その集合に属する整数の個数が答えである
    let mut r = Range {
        start: v[0].0,
        end: v[0].1,
    };
    for i in 1..m {
        // 今の区間がすべて次の区間に包含される場合
        if v[i].0 <= r.start && r.end <= v[i].1 {
            continue;
        }
        // 次の区間によって今の区間の一部が切り取られる場合
        if r.start <= v[i].0 && r.end <= v[i].1 {
            r.start = v[i].0;
            continue;
        }
        if v[i].0 <= r.start && v[i].1 <= r.end {
            r.end = v[i].1;
            continue;
        }
        // 共通部分がない場合
        println!("0");
        return;
    }

    println!("{}", r.end - r.start + 1);
}