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)

//! ----------------------------------------------
//! 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
// -----------------------------------------------

pub fn pow(x: i64, n: i64) -> i64 {
    let (mut x, mut y, mut n) = (x % P, 1_i64, n);
    while n > 0 {
        if n % 2 != 0 {
            y = (y * x) % P;
            n -= 1;
        }

        x = (x * x) % P;
        n /= 2;
    }
    y
}

/// Represents an element of finite field.
#[derive(PartialEq, Eq, PartialOrd, Ord, Clone, Copy, Default)]
struct ModInt(i64);

impl ModInt {
    fn pow(self, e: i64) -> Self {
        pow(self.0, e).into()
    }
}

impl Debug for ModInt {
    fn fmt(&self, formatter: &mut Formatter) -> std::fmt::Result {
        formatter.write_fmt(format_args!("{:?}", self.0))
    }
}

impl Display for ModInt {
    fn fmt(&self, formatter: &mut Formatter) -> std::fmt::Result {
        formatter.write_fmt(format_args!("{}", self.0))
    }
}

impl From<i64> for ModInt {
    fn from(value: i64) -> Self {
        ModInt((value % P + P) % P)
    }
}

// Generate binary operation traits.
macro_rules! impl_binary_op_for_modint {
    ($op_trait:ident, $op:ident, $assign_trait:ident, $assign:ident $(, $f:ident)*) => {
        $(impl $op_trait<ModInt> for ModInt {
            type Output = Self;

            fn $op(self, other: Self) -> Self {
                ModInt::from((self.0).$f(other.0))
            }
        })*

        impl $op_trait<i64> for ModInt {
            type Output = Self;

            fn $op(self, other: i64) -> Self {
                self.$op(ModInt::from(other))
            }
        }

        impl $assign_trait<ModInt> for ModInt {
            fn $assign(&mut self, other: Self) {
                *self = self.$op(other)
            }
        }

        impl $assign_trait<i64> for ModInt {
            fn $assign(&mut self, other: i64) {
                *self = self.$op(other)
            }
        }
    };
}

impl_binary_op_for_modint! {Add, add, AddAssign, add_assign, add}
impl_binary_op_for_modint! {Sub, sub, SubAssign, sub_assign, sub}
impl_binary_op_for_modint! {Mul, mul, MulAssign, mul_assign, mul}
impl_binary_op_for_modint! {Div, div, DivAssign, div_assign}

impl Div<ModInt> for ModInt {
    type Output = ModInt;

    fn div(self, other: Self) -> Self {
        self * other.pow(P - 2)
    }
}

const P: i64 = 1_000_000_007;

fn main() {
    let (sn, tn) = read!(usize, usize);
    let S = read![[i64]];
    let T = read![[i64]];

    let mut inc = vec![ModInt::from(0); sn + 1];

    // dp[si] = n : S[..si] の部分列と T[..ti] の部分列の組み合わせの総数
    let mut dp = vec![ModInt::from(1); sn + 1];

    for ti in 0..tn {
        for z in 0..sn {
            if S[z] == T[ti] {
                inc[z] += dp[z];
            }
        }

        for si in 0..sn {
            dp[si + 1] = dp[si] + inc[si];
        }

        debug!(ti, inc, dp);
    }

    println!("{}", dp[sn])
}

/*

4 3
1 2 3 1
2 1 3

#=> 10

*/