/*

HDU 6078 - Wavel Sequence [ DP ]  |  2017 Multi-University Training Contest 4

题意：

	给定 a[N], b[M]

	要求满足

			a[f(1)]<a[f(2)]>a[f(3)]<a[f(4)]>a[f(5)]<a[f(6)]...

			b[g(i)] == a[f(i)]

			f(i) < f(i+1), g(i) < g(i+1)

	的子序列 的数目

分析：

	dp[i][j][0] 表示 以a[i], b[j]结尾的且为波谷的情况总和，dp[i][j][1] 为波峰

	对于某个 i,j满足 a[i] == b[j]

	dp[i][j][0] = sum(dp[x][y][1]), x < i && y < j && b[y] > a[i]

	设 sum[i-1][y][1] = ∑dp[x][y][1] , x <= i-1

	则 dp[i][j][0] = ∑ sum[i-1][y][1], b[y] > a[i]

	对于每一个 b[j]， sum[i][j] 累计了前i个的a[i]的影响，求出每次dp[i][j]后O(1)更新即可

	即	sum[i][j][1] = sum[i-1][j][1] + dp[i][j][1]

	然后对于某一个a[i]的所有b[j]，可以以前缀和的形式，利用sum[i-1][y]，O(n)全部更新

	所以复杂度O(n^2)

*/

#include <bits/stdc++.h>

using namespace std;

const int MOD = 998244353;

const int N = 2005;

int t, n, m;

int a[N], b[N];

int dp[N][2];

int sum[N][2];

int main()

{

    scanf("%d", &t);

    while (t--)

    {

        scanf("%d%d", &n, &m);

        for (int i = 1; i <= n; i++) scanf("%d", &a[i]);

        for (int i = 1; i <= m; i++) scanf("%d", &b[i]);

        memset(sum, 0, sizeof(sum));

        long long ans = 0;

        for (int i = 1; i <= n; i++)

        {

            int cnt0 = 0, cnt1 = 0;

            for (int j = 1; j <= m; j++)

            {

                if (a[i] == b[j])

                {

                    dp[j][0] = cnt1 + 1;

                    dp[j][1] = cnt0;

                    ans = (ans + cnt1+cnt0+1) % MOD;

                }

                else if (b[j] > a[i])

                    cnt1 = (cnt1+ sum[j][1]) % MOD;

                else

                    cnt0 = (cnt0+ sum[j][0]) % MOD;

            }

            for (int j = 1; j <= m; j++)

                if (a[i] == b[j])

                {

                    sum[j][0] = (sum[j][0] + dp[j][0]) % MOD;

                    sum[j][1] = (sum[j][1] + dp[j][1]) % MOD;

                }

        }

        printf("%lld\n", ans);

    }

}