Problem:
Score : 700 points

Problem Statement

There are N stones arranged in a row. The i-th stone from the left is painted in the color C_i.

Snuke will perform the following operation zero or more times:

Choose two stones painted in the same color. Repaint all the stones between them, with the color of the chosen stones.

Find the number of possible final sequences of colors of the stones, modulo 10^9+7.

Constraints

1 \leq N \leq 2\times 10^5

1 \leq C_i \leq 2\times 10^5(1\leq i\leq N)

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N
C_1
:
C_N

Output

Print the number of possible final sequences of colors of the stones, modulo 10^9+7.

Sample Input 1

5
1
2
1
2
2

Sample Output 1

3

We can make three sequences of colors of stones, as follows:

(1,2,1,2,2), by doing nothing.

(1,1,1,2,2), by choosing the first and third stones to perform the operation.

(1,2,2,2,2), by choosing the second and fourth stones to perform the operation.

Sample Input 2

6
4
2
5
4
2
4

Sample Output 2

5

Sample Input 3

7
1
3
1
2
3
3
2

Sample Output 3

5

Solution in Rust (1.15.1) (judge verdict: Runtime Error)

//! ----------------------------------------------
//! Framework <https://github.com/vain0x/procon>
//!
//! See the bottom of file for solution.
//! ----------------------------------------------

#![allow(unused_imports)]
#![allow(non_snake_case)]

use std::cell::RefCell;
use std::cmp::{max, min, Ordering};
use std::collections::*;
use std::fmt::{Debug, Display, Formatter, Write as FmtWrite};
use std::io::{stderr, stdin, BufRead, Write};
use std::mem::{replace, swap};
use std::ops::*;
use std::rc::Rc;

/// Print values to standard error if debug mode.
#[allow(unused_macros)]
macro_rules! debug {
    ($($e:expr),*) => {
        #[cfg(debug_assertions)]
        $({
            let (e, mut err) = (stringify!($e), stderr());
            writeln!(err, "\x1B[33m{}\x1B[0m = {:?}", e, $e).unwrap()
        })*
    };
}

/// Read from standard input and parse each word.
/// - `read!(T, U, ..)` parses a line as a tuple of words.
/// - `read![[T]]` parses a line as an array of words.
/// - `read![..; N]` parses `N` lines, using `read!(..)` repeatedly.
#[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),*) => {{
        let buf = rl();
        let mut w = buf.split_whitespace();
        ($(w.next().unwrap().parse::<$t>().unwrap()),*)
    }};
}

/// Read a line from standard input.
#[allow(dead_code)]
fn rl() -> String {
    let mut buf = String::new();
    stdin().read_line(&mut buf).unwrap();

    #[allow(deprecated)]
    buf.trim_right().to_owned()
}

// -----------------------------------------------
// Solution
// -----------------------------------------------

const P: i64 = 1_000_000_007;

struct Solver {
    N: usize,
    C: Vec<usize>,
    next: Vec<usize>,
    dp: Vec<i64>,
}

impl Solver {
    fn dfs(&mut self, i: usize) -> i64 {
        if self.dp[i] == P {
            self.dp[i] = self.do_dfs(i);
            // debug!(i, self.dp[i]);
        }
        self.dp[i]
    }

    fn do_dfs(&mut self, i: usize) -> i64 {
        if i >= self.N {
            return 1;
        }

        let mut sum = 0;

        let mut r = i;
        while r < self.N {
            // i..=r を C[i] で塗った
            sum += self.dfs(r + 1);
            sum %= P;

            r = self.next[r];
        }

        sum
    }
}

fn solve() {
    let N = read!(usize);
    let mut C = read![usize; N];

    for i in 0..N {
        C[i] -= 1;
    }

    let mut prev = vec![N; N + 1];
    let mut next = vec![N; N + 1];
    let mut i = 0;
    while i < N {
        while i + 1 < N && C[i] == C[i + 1] {
            i += 1;
        }
        next[prev[C[i]]] = i;
        prev[C[i]] = i;
        i += 1;
    }
    // debug!(next);

    // dp[r]: 0..r の塗り方が決まって、r.. に操作が及んでないときの場合の数

    let dp = vec![P; N + 1];

    println!(
        "{}",
        Solver {
            N: N,
            C: C,
            next: next,
            dp: dp
        }
        .dfs(0)
    );
}

fn main() {
    std::thread::Builder::new()
        .stack_size(256 * 1024 * 1024)
        .spawn(|| solve())
        .unwrap()
        .join()
        .unwrap();
}