Problem:
Score : 500 points

Problem Statement

You are given two integer sequences S and T of length N and M, respectively, both consisting of integers between 1 and 10^5 (inclusive).

In how many pairs of a subsequence of S and a subsequence of T do the two subsequences are the same in content?

Here the subsequence of A is a sequence obtained by removing zero or more elements from A and concatenating the remaining elements without changing the order.

For both S and T, we distinguish two subsequences if the sets of the indices of the removed elements are different, even if the subsequences are the same in content.

Since the answer can be tremendous, print the number modulo 10^9+7.

Constraints

1 \leq N, M \leq 2 \times 10^3

The length of S is N.

The length of T is M.

1 \leq S_i, T_i \leq 10^5

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N M
S_1 S_2 ... S_{N-1} S_{N}
T_1 T_2 ... T_{M-1} T_{M}

Output

Print the number of pairs of a subsequence of S and a subsequence of T such that the subsequences are the same in content, modulo 10^9+7.

Sample Input 1

2 2
1 3
3 1

Sample Output 1

3

S has four subsequences: (), (1), (3), (1, 3).

T has four subsequences: (), (3), (1), (3, 1).

There are 1 \times 1 pair of subsequences in which the subsequences are both (), 1 \times 1 pair of subsequences in which the subsequences are both (1), and 1 \times 1 pair of subsequences in which the subsequences are both (3), for a total of three pairs.

Sample Input 2

2 2
1 1
1 1

Sample Output 2

6

S has four subsequences: (), (1), (1), (1, 1).

T has four subsequences: (), (1), (1), (1, 1).

There are 1 \times 1 pair of subsequences in which the subsequences are both (), 2 \times 2 pairs of subsequences in which the subsequences are both (1), and 1 \times 1 pair of subsequences in which the subsequences are both (1,1), for a total of six pairs.
Note again that we distinguish two subsequences if the sets of the indices of the removed elements are different, even if the subsequences are the same in content.

Sample Input 3

4 4
3 4 5 6
3 4 5 6

Sample Output 3

16

Sample Input 4

10 9
9 6 5 7 5 9 8 5 6 7
8 6 8 5 5 7 9 9 7

Sample Output 4

191

Sample Input 5

20 20
1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1

Sample Output 5

846527861

Be sure to print the number modulo 10^9+7.

Solution in Rust (1.15.1) (judge verdict: Time Limit Exceeded)

use std::fmt::Display;
use std::io::{stdin, Read};
use std::ops::{Add, AddAssign, Div, DivAssign, Mul, MulAssign, Sub, SubAssign};
use std::str::FromStr;
pub const MOD: usize = 1000_000_007;

#[derive(Clone, Copy, PartialEq, Debug, Default)]
pub struct Mint<T>(pub T);

impl Mint<usize> {
    pub fn pow(self, exp: usize) -> Self {
        let mut exp = exp;
        let mut v = self;
        let mut ret = Mint(1);
        while exp > 0 {
            if exp & 1 == 1 {
                ret *= v;
            }
            v *= v;
            exp >>= 1;
        }
        ret
    }
    pub fn inv(self) -> Self {
        self.pow(MOD - 2)
    }
}

impl<T: Display + Copy + PartialEq> Display for Mint<T> {
    fn fmt(&self, f: &mut std::fmt::Formatter) -> std::fmt::Result {
        write!(f, "{}", self.0)
    }
}

impl<T: FromStr + Copy + PartialEq> FromStr for Mint<T> {
    type Err = T::Err;
    fn from_str(s: &str) -> Result<Self, Self::Err> {
        let v = T::from_str(s)?;
        Ok(Mint(v))
    }
}

impl Add<usize> for Mint<usize> {
    type Output = Self;
    fn add(self, rhs: usize) -> Self::Output {
        Mint((self.0 + rhs) % MOD)
    }
}

impl Add for Mint<usize> {
    type Output = Self;
    fn add(self, rhs: Self) -> Self::Output {
        self + rhs.0
    }
}

impl AddAssign<usize> for Mint<usize> {
    fn add_assign(&mut self, rhs: usize) {
        *self = *self + rhs;
    }
}

impl AddAssign for Mint<usize> {
    fn add_assign(&mut self, rhs: Self) {
        *self = *self + rhs;
    }
}

impl Sub<usize> for Mint<usize> {
    type Output = Self;
    fn sub(self, rhs: usize) -> Self::Output {
        let rhs = rhs % MOD;
        let lhs = self.0 + if self.0 < rhs { MOD } else { 0 };
        Mint(lhs - rhs)
    }
}

impl Sub for Mint<usize> {
    type Output = Self;
    fn sub(self, rhs: Self) -> Self::Output {
        self - rhs.0
    }
}

impl SubAssign<usize> for Mint<usize> {
    fn sub_assign(&mut self, rhs: usize) {
        *self = *self - rhs
    }
}

impl SubAssign for Mint<usize> {
    fn sub_assign(&mut self, rhs: Self) {
        *self = *self - rhs
    }
}

impl Mul<usize> for Mint<usize> {
    type Output = Self;
    fn mul(self, rhs: usize) -> Self::Output {
        Mint((self.0 * rhs) % MOD)
    }
}

impl Mul for Mint<usize> {
    type Output = Self;
    fn mul(self, rhs: Self) -> Self::Output {
        self * rhs.0
    }
}

impl MulAssign<usize> for Mint<usize> {
    fn mul_assign(&mut self, rhs: usize) {
        *self = *self * rhs;
    }
}

impl MulAssign for Mint<usize> {
    fn mul_assign(&mut self, rhs: Self) {
        *self = *self * rhs;
    }
}

impl Div<usize> for Mint<usize> {
    type Output = Self;
    fn div(self, rhs: usize) -> Self::Output {
        self * Mint(rhs).pow(MOD - 2)
    }
}

impl Div for Mint<usize> {
    type Output = Self;
    fn div(self, rhs: Self) -> Self::Output {
        self / rhs.0
    }
}

impl DivAssign<usize> for Mint<usize> {
    fn div_assign(&mut self, rhs: usize) {
        *self = *self / rhs;
    }
}
impl DivAssign for Mint<usize> {
    fn div_assign(&mut self, rhs: Self) {
        *self = *self / rhs;
    }
}
macro_rules! parse {
    ($it: ident ) => {};
    ($it: ident, ) => {};
    ($it: ident, $var:ident : $t:tt $($r:tt)*) => {
        let $var = parse_val!($it, $t);
        parse!($it $($r)*);
    };
    ($it: ident, mut $var:ident : $t:tt $($r:tt)*) => {
        let mut $var = parse_val!($it, $t);
        parse!($it $($r)*);
    };
}

macro_rules! parse_val {
    ($it: ident, [$t:tt; $len:expr]) => {
        (0..$len).map(|_| parse_val!($it, $t)).collect::<Vec<_>>();
    };
    ($it: ident, $t: ty) => {
        $it.next().unwrap().parse::<$t>().unwrap()
    };
}

fn solve(s: &str) {
    let mut it = s.split_whitespace();
    parse! {
        it,
        n: usize,
        m: usize,
        s:[usize;n],
        t:[usize;m]
    };
    let mut dp = vec![vec![Mint(0); m + 1]; n + 1];
    let mut sd = std::collections::HashMap::new();
    let mut td = std::collections::HashMap::new();
    for i in 0..n {
        sd.entry(s[i]).or_insert(Vec::new()).push(i);
    }
    for i in 0..m {
        td.entry(t[i]).or_insert(Vec::new()).push(i);
    }
    for i in 0..n {
        for j in 0..m {
            let x = dp[i + 1][j];
            dp[i + 1][j + 1] += x;
            if let Some(v) = sd.get(&t[j]) {
                for &k in v.iter() {
                    if k > i {
                        break;
                    }
                    let x = dp[k][j];
                    dp[i + 1][j + 1] += x + 1;
                }
            }
        }
    }
    println!("{}", dp[n][m] + 1); // DEBUG
}

fn main() {
    let mut s = String::new();
    stdin().read_to_string(&mut s).unwrap();
    solve(&s);
}

#[cfg(test)]
mod tests {
    use super::*;
    #[test]
    fn test_input() {
        let s = "
        ";
        solve(s);
    }
}