Break This Array

#include<bits/stdc++.h>
using namespace std;

#define INF 2047483647

#define ll long long
#define MOD 1000000007

const int N = 505;
ll dp[N][N][N];

ll inverse[N];
void solve()
{
    int n,K;
    cin>>n>>K;

    for(int i=0;i<=n;i++)
    {
        for(int j=0;j<=n;j++)
        {
            for(int k=0;k<=K;k++)
                dp[i][j][k] = 0;
        }
    }

    ll a[n];
    for(int i=0;i<n;i++)
        cin>>a[i];

    string s;
    cin>>s;

    s="#"+s+"L";
    dp[1][n][0] = 1;

    for(int k=1;k<=K+1;k++)
    {
        for(int len=n;len>0;len--)
        {
            for(int i=1;i<=n;i++)
            {
                int j = i+len-1;

                if( j>n ) continue;

                if( k>1 and s[k-1] == 'L' )
                {
                    dp[i][j][k-1]+=dp[i][j+1][k-1];
                    dp[i][j][k-1]%=MOD;
                }
                else if(k>1 and s[k-1] == 'R')
                {
                    dp[i][j][k-1]+=dp[i-1][j][k-1];
                    dp[i][j][k-1]%=MOD;
                }
                if( i == j )
                {
                    dp[i][j][k]+=dp[i][j][k-1];
                    dp[i][j][k]%=MOD;
                    continue;
                }
                ll div = inverse[j-i];
                div = div*dp[i][j][k-1];
                div%=MOD;
                if( s[k] == 'L' )
                {
                    dp[i][j-1][k]+=div;
                    dp[i][j-1][k]%=MOD;
                }
                else
                {
                    dp[i+1][j][k]+=div;
                    dp[i+1][j][k]%=MOD;
                }
            }
        }
    }


    ll sum[n+1] = {0};
    for(int i=1;i<=n;i++)
        sum[i] = (sum[i-1]+a[i-1])%MOD;

    ll res = 0;
    for(int i =1;i<=n;i++)
    {
        for(int j=i;j<=n;j++)
        {
            ll x = dp[i][j][K]*(sum[j]-sum[i-1]+MOD);
            x%=MOD;
            (res+=x)%=MOD;
        }
    }

    cout<<res<<endl;

}


int main(){
    function<ll int(ll int)> InverseUnderModulo = [&](ll int x)
    {
        if(x == 1)
            return 1ll;

        return (MOD - ((MOD/x) * InverseUnderModulo(MOD%x)) % MOD) % MOD;
    };

    for(int i=1;i<N;i++) inverse[i] = InverseUnderModulo(i);
    int tc=1;
    cin>>tc;
    while(tc--)
        solve();
}