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 Java8 (OpenJDK 1.8.0) (judge verdict: Wrong Answer)

import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Collections;
import java.util.HashMap;
import java.util.List;
import java.util.Map;
import java.util.NoSuchElementException;
import java.util.function.BiFunction;
import java.util.function.Consumer;
import java.util.function.Supplier;

public class Main {
    final static int MOD07 = 1000000007;
    final static long MOD09 = 1_000_000_009;

    public static void main(String[] args) {
        final FastScanner fs = new FastScanner();
        PrintWriter out = new PrintWriter(System.out);
        int N = fs.nextInt();
        int M = fs.nextInt();
        int[] S = new int[N];
        int[] T = new int[M];
        for (int i = 0; i < N; i++) {
            S[i] = fs.nextInt();
        }
        for (int i = 0; i < M; i++) {
            T[i] = fs.nextInt();
        }

        int[][] dp = new int[N+1][M+1]; // dp_i_j
        int[][] sdp = new int[N+2][M+2]; // dp_i_j
        for (int i = 0; i <= N; i++) {
            Arrays.fill(dp[i], 0);
            Arrays.fill(sdp[i], 0);
        }
        Arrays.fill(sdp[N+1], 0);
        dp[0][0] = 1;
        sdp[1][1] = 1;

        for (int i = 0; i <= N; ++i) {
            for (int j = 0; j <= M; ++j) {
                if (i == 0 && j == 0) continue;
                if (i-1 >= 0 && j-1 >= 0 && S[i-1] == T[j-1]) {
                    dp[i][j] = sdp[i][j];
                }
                int val = sdp[i+1][j] + sdp[i][j+1];
                val %= MOD07;
                val -= sdp[i][j];
                val %= MOD07;
                val += dp[i][j];
                val %= MOD07;
                sdp[i+1][j+1] = val;
            }
        }

        out.println(sdp[N+1][M+1]);
        out.flush();
    }

    static void debug(Object... os) {
        System.err.println(Arrays.deepToString(os));
    }

    // Execute func for all patterns can be represented with n bit.
    // O(n2^n)
    static void forAllBit(int n, Consumer<List<Boolean>> func) {
        for (int i = 0; i < (1 << n); i++) {
            List<Boolean> arg = new ArrayList();
            for (int j = 0; j < n; j++) {
                boolean isJthBitSet = ((i >> j) & 1) == 1;
                arg.add(isJthBitSet);
            }
            func.accept(arg);
        }
    }

    static class Pair<F extends Comparable<F>, S extends Comparable<S>> implements Comparable<Pair<F, S>> {
        F f;
        S s;

        Pair() {
        }

        Pair(F f, S s) {
            this.f = f;
            this.s = s;
        }

        Pair(Pair<F, S> p) {
            f = p.f;
            s = p.s;
        }

        @Override
        public int compareTo(Pair<F, S> p) {
            if (f.compareTo(p.f) != 0) {
                return f.compareTo(p.f);
            }
            return s.compareTo(p.s);
        }

        @Override
        public int hashCode() {
            return f.hashCode() ^ s.hashCode();
        }

        @Override
        public boolean equals(Object o) {
            if (this == o) {
                return true;
            }
            if (o == null || this.f == null || this.s == null) {
                return false;
            }
            if (this.getClass() != o.getClass()) {
                return false;
            }
            Pair p = (Pair) o;
            return this.f.equals(p.f) && this.s.equals(p.s);
        }

        @Override
        public String toString() {
            return "{" + f.toString() + ", " + s.toString() + "}";
        }
    }

    public static int gcd(int a, int b) {
        if (b == 0) {
            return a;
        }
        return gcd(b, a % b);
    }

    public static boolean isPrime(int n) {
        for (int i = 2; i <= ((int) Math.sqrt(n)); i++) {
            if (n % i == 0) {
                return false;
            }
        }
        return n != 1;
    }

    public static void reverse(int[] array) {
        for (int i = 0; i < array.length / 2; i++) {
            int temp = array[i];
            array[i] = array[array.length - i - 1];
            array[array.length - i - 1] = temp;
        }
    }

    public static void reverse(long[] array) {
        for (int i = 0; i < array.length / 2; i++) {
            long temp = array[i];
            array[i] = array[array.length - i - 1];
            array[array.length - i - 1] = temp;
        }
    }

    public static boolean isPrime(long n) {
        for (long i = 2; i <= ((long) Math.sqrt(n)); i++) {
            if (n % i == 0) {
                return false;
            }
        }
        return n != 1;
    }

    public static Map<Long, Long> primeFactor(long n) {
        Map<Long, Long> map = new HashMap<>();
        long num = n;
        for (long i = 2; i <= ((long) Math.sqrt(n)); i++) {
            while (num % i == 0L) {
                map.put(i, map.getOrDefault(i, 0L) + 1L);
                num /= i;
            }
        }
        if (num != 1) {
            map.put(num, 1L);
        }

        return map;
    }

    public static List<Long> divisors(long n) {
        List<Long> list = new ArrayList<>();
        for (long i = 1; i <= ((long) Math.sqrt(n)); i++) {
            if (n % i == 0) {
                list.add(i);
                if (i != n / i) {
                    list.add(n / i);
                }
            }
        }

        return list;
    }

    public static int lcm(int a, int b) {
        return a / gcd(a, b) * b;
    }

    public static int nCk(int n, int k, int M) {
        long ret = 1;
        int min = Math.min(k, n - k);
        for (int i = 1; i <= min; i++) {
            ret = (ret * pow(i, M - 2, M)) % M;
        }
        for (int i = n - min + 1; i <= n; i++) {
            ret = (ret * i) % M;
        }
        return (int) ret;
    }

    //return a^b mod M O(logB)
    public static long pow(long a, long b, int M) {
        long ret = 1;
        long tmp = a;
        while (b > 0) {
            if ((b & 1) == 1) ret = (ret * tmp) % M;
            tmp = (tmp * tmp) % M;
            b = b >> 1;
        }
        return ret;
    }


    public static int find(int[] tree, int idx) {
        if (tree[idx] == idx) return idx;
        else return tree[idx] = find(tree, tree[idx]);
    }

    public static void union(int[] tree, int idx1, int idx2) {
        int root1 = find(tree, idx1);
        int root2 = find(tree, idx2);
        tree[root2] = root1;
    }

    public static int lowerBound(int[] list, int target) {
        int ok = list.length;
        int ng = -1;
        while (Math.abs(ok - ng) > 1) {
            int mid = (ok + ng) / 2;
            if (list[mid] >= target) {
                ok = mid;
            } else {
                ng = mid;
            }
        }
        return ok;
    }

    public static int lowerBound(long[] list, long target) {
        int ok = list.length;
        int ng = -1;
        while (Math.abs(ok - ng) > 1) {
            int mid = (ok + ng) / 2;
            if (list[mid] >= target) {
                ok = mid;
            } else {
                ng = mid;
            }
        }
        return ok;
    }

    public static <T extends Comparable<? super T>> int lowerBound(List<T> list, T target) {
        return ~Collections.binarySearch(list, target, (x, y) -> (x.compareTo(y) >= 0) ? 1 : -1);
    }

    public static <T extends Comparable<? super T>> int upperBound(List<T> list, T target) {
        return ~Collections.binarySearch(list, target, (x, y) -> (x.compareTo(y) > 0) ? 1 : -1);
    }
}


class FastScanner {
    private final InputStream in = System.in;
    private final byte[] buffer = new byte[1024];
    private int ptr = 0;
    private int buflen = 0;

    private boolean hasNextByte() {
        if (ptr < buflen) {
            return true;
        } else {
            ptr = 0;
            try {
                buflen = in.read(buffer);
            } catch (IOException e) {
                e.printStackTrace();
            }
            if (buflen <= 0) {
                return false;
            }
        }
        return true;
    }

    private int readByte() {
        if (hasNextByte()) return buffer[ptr++];
        else return -1;
    }

    private boolean isPrintableChar(int c) {
        return 33 <= c && c <= 126;
    }

    public boolean hasNext() {
        while (hasNextByte() && !isPrintableChar(buffer[ptr])) ptr++;
        return hasNextByte();
    }

    public String next() {
        if (!hasNext()) throw new NoSuchElementException();
        StringBuilder sb = new StringBuilder();
        int b = readByte();
        while (isPrintableChar(b)) {
            sb.appendCodePoint(b);
            b = readByte();
        }
        return sb.toString();
    }

    public long nextLong() {
        if (!hasNext()) throw new NoSuchElementException();
        long n = 0;
        boolean minus = false;
        int b = readByte();
        if (b == '-') {
            minus = true;
            b = readByte();
        }
        if (b < '0' || '9' < b) {
            throw new NumberFormatException();
        }
        while (true) {
            if ('0' <= b && b <= '9') {
                n *= 10;
                n += b - '0';
            } else if (b == -1 || !isPrintableChar(b)) {
                return minus ? -n : n;
            } else {
                throw new NumberFormatException();
            }
            b = readByte();
        }
    }

    public int nextInt() {
        long nl = nextLong();
        if (nl < Integer.MIN_VALUE || nl > Integer.MAX_VALUE) throw new NumberFormatException();
        return (int) nl;
    }

    public double nextDouble() {
        return Double.parseDouble(next());
    }
}

class BIT<T> {
    int n;
    ArrayList<T> bit;
    BiFunction<T, T, T> bif;

    /**
     * 1-indexed なBinary Indexed Treeを構築する
     *
     * @param n   容量
     * @param bif 適用させる関数
     * @param sup 初期値
     */
    BIT(int n, BiFunction<T, T, T> bif, Supplier<T> sup) {
        this.n = n;
        bit = new ArrayList<>(n + 1);
        for (int i = 0; i < n + 1; ++i) {
            bit.add(sup.get());
        }
        this.bif = bif;
    }

    /**
     * iの位置の値をvで更新する
     *
     * @param i index
     * @param v 新しい値
     */
    void set(int i, T v) {
        for (int x = i; x <= n; x += x & -x) {
            bit.set(x, bif.apply(bit.get(x), v));
        }
    }

    /**
     * クエリー
     *
     * @param defaultValue 初期値
     * @param i            index
     * @return [1, i]までfを適用した結果
     */
    T reduce(T defaultValue, int i) {
        T ret = defaultValue;
        for (int x = i; x > 0; x -= x & -x) {
            ret = bif.apply(ret, bit.get(x));
        }
        return ret;
    }
}

class SegmentTree<T> {
    int n;
    ArrayList<T> dat;
    BiFunction<T, T, T> bif;
    Supplier<T> sup;

    /**
     * 0-indexed なSegment Treeを構築する
     *
     * @param n_  要求容量
     * @param bif 適用させる関数
     * @param sup 初期値
     */
    SegmentTree(int n_, BiFunction<T, T, T> bif, Supplier<T> sup) {
        n = 1;
        while (n < n_) n *= 2;

        dat = new ArrayList<>(2 * n - 1);
        for (int i = 0; i < 2 * n - 1; ++i) {
            dat.add(sup.get());
        }
        this.bif = bif;
        this.sup = sup;
    }

    /**
     * kの位置の値をvで更新する
     *
     * @param k index
     * @param v 新しい値
     */
    void set(int k, T v) {
        k += n - 1;
        dat.set(k, v);
        while (k > 0) {
            k = (k - 1) / 2;
            dat.set(k, bif.apply(dat.get(k * 2 + 1), dat.get(k * 2 + 2)));
        }
    }

    /**
     * クエリー
     *
     * @param l はじめ
     * @param r おわり
     * @return [l, r)での演算bifを適用した結果を返す
     */
    T reduce(int l, int r) {
        return _reduce(l, r, 0, 0, n);
    }

    T _reduce(int a, int b, int k, int l, int r) {
        if (r <= a || b <= l) return sup.get();
        if (a <= l && r <= b) return dat.get(k);
        T vl = _reduce(a, b, k * 2 + 1, l, (l + r) / 2);
        T vr = _reduce(a, b, k * 2 + 2, (l + r) / 2, r);
        return bif.apply(vl, vr);
    }
}

class UnionFind {
    int[] par;

    UnionFind(int n) {
        par = new int[n];
        for (int i = 0; i < n; ++i) {
            par[i] = i;
        }
    }

    int find(int x) {
        if (par[x] == x) {
            return x;
        }
        return par[x] = find(par[x]);
    }

    boolean same(int x, int y) {
        return find(x) == find(y);
    }

    void union(int x, int y) {
        x = find(x);
        y = find(y);
        if (x == y) {
            return;
        }
        par[x] = y;
    }
}