Problem:
Score : 1200 points

Problem Statement

Takahashi is an expert of Clone Jutsu, a secret art that creates copies of his body.

On a number line, there are N copies of Takahashi, numbered 1 through N.
The i-th copy is located at position X_i and starts walking with velocity V_i in the positive direction at time 0.

Kenus is a master of Transformation Jutsu, and not only can he change into a person other than himself, but he can also transform another person into someone else.

Kenus can select some of the copies of Takahashi at time 0, and transform them into copies of Aoki, another Ninja.
The walking velocity of a copy does not change when it transforms.
From then on, whenever a copy of Takahashi and a copy of Aoki are at the same coordinate, that copy of Takahashi transforms into a copy of Aoki.

Among the 2^N ways to transform some copies of Takahashi into copies of Aoki at time 0, in how many ways will all the copies of Takahashi become copies of Aoki after a sufficiently long time?
Find the count modulo 10^9+7.

Constraints

1 ≤ N ≤ 200000

1 ≤ X_i,V_i ≤ 10^9(1 ≤ i ≤ N)

X_i and V_i are integers.

All X_i are distinct.

All V_i are distinct.

Input

The input is given from Standard Input in the following format:

N
X_1 V_1
:
X_N V_N

Output

Print the number of the ways that cause all the copies of Takahashi to turn into copies of Aoki after a sufficiently long time, modulo 10^9+7.

Sample Input 1

3
2 5
6 1
3 7

Sample Output 1

6

All copies of Takahashi will turn into copies of Aoki after a sufficiently long time if Kenus transforms one of the following sets of copies of Takahashi into copies of Aoki: (2), (3), (1,2), (1,3), (2,3) and (1,2,3).

Sample Input 2

4
3 7
2 9
8 16
10 8

Sample Output 2

9

Solution in Rust (1.15.1) (judge verdict: Accepted)

#[allow(unused_imports)]
use std::cmp::*;
#[allow(unused_imports)]
use std::collections::*;
use std::io::Read;
#[allow(dead_code)]
fn getline() -> String {
    let mut ret = String::new();
    std::io::stdin().read_line(&mut ret).ok().unwrap();
    ret
}
fn get_word() -> String {
    let mut stdin = std::io::stdin();
    let mut u8b: [u8; 1] = [0];
    loop {
        let mut buf: Vec<u8> = Vec::with_capacity(16);
        loop {
            let res = stdin.read(&mut u8b);
            if res.unwrap_or(0) == 0 || u8b[0] <= b' ' {
                break;
            } else {
                buf.push(u8b[0]);
            }
        }
        if buf.len() >= 1 {
            let ret = String::from_utf8(buf).unwrap();
            return ret;
        }
    }
}

#[allow(dead_code)]
fn get<T: std::str::FromStr>() -> T { get_word().parse().ok().unwrap() }

/// Coordinate compression
/// Returns a vector of usize, with i-th element the "rank" of a[i] in a.
/// The property forall i. inv_map[ret[i]] == a[i] holds.
/// Verified by: yukicoder No.404 (http://yukicoder.me/submissions/155377)
fn coord_compress<T: Ord>(a: &[T])
                          -> (Vec<usize>, Vec<&T>) {
    let n = a.len();
    let mut cp: Vec<(&T, usize)> = (0 .. n).map(|i| (&a[i], i)).collect();
    cp.sort();
    let mut inv_map = Vec::with_capacity(n);
    let mut prev: Option<&T> = None;
    let mut ret = vec![0; n];
    let mut cnt = 0;
    for (v, i) in cp {
        if prev == Some(v) {
            ret[i] = cnt - 1;
            continue;
        }
        ret[i] = cnt;
        inv_map.push(v);
        prev = Some(v);
        cnt += 1;
    }
    (ret, inv_map)
}


/**
 * Segment Tree. This data structure is useful for fast folding on intervals of an array
 * whose elements are elements of monoid M. Note that constructing this tree requires the identity
 * element of M and the operation of M.
 * Verified by: yukicoder No. 259 (http://yukicoder.me/submissions/100581)
 */
struct SegTree<I, BiOp> {
    n: usize,
    dat: Vec<I>,
    op: BiOp,
    e: I,
}

impl<I, BiOp> SegTree<I, BiOp>
    where BiOp: Fn(I, I) -> I,
          I: Copy {
    pub fn new(n_: usize, op: BiOp, e: I) -> Self {
        let mut n = 1;
        while n < n_ { n *= 2; } // n is a power of 2
        SegTree {n: n, dat: vec![e; 2 * n - 1], op: op, e: e}
    }
    /* ary[k] <- v */
    pub fn update(&mut self, idx: usize, v: I) {
        let mut k = idx + self.n - 1;
        self.dat[k] = v;
        while k > 0 {
            k = (k - 1) / 2;
            self.dat[k] = (self.op)(self.dat[2 * k + 1], self.dat[2 * k + 2]);
        }
    }
    /* [a, b) (note: half-inclusive)
     * http://proc-cpuinfo.fixstars.com/2017/07/optimize-segment-tree/ */
    pub fn query(&self, mut a: usize, mut b: usize) -> I {
        let mut y = self.e;
        a += self.n - 1;
        b += self.n - 1;
        while a < b {
            if (a & 1) == 0 {
                y = (self.op)(self.dat[a], y);
            }
            if (b & 1) == 0 {
                y = (self.op)(y, self.dat[b - 1]);
            }
            a = a / 2;
            b = (b - 1) / 2;
        }
        y
    }
}

const MOD: i64 = 1_000_000_007;

fn calc(range: Vec<(usize, usize)>, lim: usize) -> i64 {
    let mut st = SegTree::new(lim + 1, |x, y| (x + y) % MOD, 0);
    st.update(0, 1);
    for (l, r) in range {
        let tmp = st.query(l - 1, r + 1) + st.query(r, r + 1);
        st.update(r, tmp);
    }
    st.query(lim, lim + 1)
}

const INF: i64 = 1 << 50;
// This solution was written after the author read the editorial.
fn solve() {
    let n = get();
    let mut xv: Vec<(i64, i64)> = Vec::new();
    for _ in 0 .. n {
        let x = get();
        let v = get();
        xv.push((x, v));
    }
    xv.sort();
    let mut range = vec![(xv[n - 1].1, xv[0].1); n];
    for i in 1 .. n {
        range[i].1 = max(range[i - 1].1, xv[i].1);
    }
    for i in (0 .. n - 1).rev() {
        range[i].0 = min(range[i + 1].0, xv[i].1);
    }
    // Coord comp
    let mut vs: Vec<i64> = xv.iter().map(|&x| x.1).collect();
    vs.push(-INF);
    vs.push(INF);
    let (_, comp) = coord_compress(&vs);
    let mut mp = HashMap::new();
    for i in 0 .. comp.len() {
        mp.insert(*comp[i], i);
    }
    let range: Vec<(usize, usize)> = range.into_iter().map(|(x, y)|
                                      (*mp.get(&x).unwrap(), *mp.get(&y).unwrap()))
        .collect();
    println!("{}", calc(range, comp.len() - 2));
}

fn main() {
    // In order to avoid potential stack overflow, spawn a new thread.
    let stack_size = 104_857_600; // 100 MB
    let thd = std::thread::Builder::new().stack_size(stack_size);
    thd.spawn(|| solve()).unwrap().join().unwrap();
}