// #pragma GCC target("avx")
// #pragma GCC optimize("O3")
// #pragma GCC optimize("unroll-loops")

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

#define rep(i,n) for(int i = 0; i < (int)n; i++)
#define FOR(n) for(int i = 0; i < (int)n; i++)
#define repi(i,a,b) for(int i = (int)a; i < (int)b; i++)
#define all(x) x.begin(),x.end()
//#define mp make_pair
#define vi vector<int>
#define vvi vector<vi>
#define vvvi vector<vvi>
#define vvvvi vector<vvvi>
#define pii pair<int,int>
#define vpii vector<pair<int,int>>

template<typename T>
bool chmax(T &a, const T b) {if(a<b) {a=b; return true;} else {return false;}}
template<typename T>
bool chmin(T &a, const T b) {if(a>b) {a=b; return true;} else {return false;}}

using ll = long long;
using ld = long double;
using ull = unsigned long long;

const ll INF = 1e10;
const ld pi = 3.1415926535897932384626433832795028;
const ll mod = 998244353;
int dx[] = {1, 0, -1, 0, -1, -1, 1, 1};
int dy[] = {0, 1, 0, -1, -1, 1, -1, 1};

#define int long long

void solve() {
    int n, m;
    cin >> n >> m;
    vi l(n), r(n);
    int l_sum = 0, r_sum = 0;
    FOR(n) {
        cin >> l[i];
        l_sum += l[i];
    }
    FOR(n) {
        cin >> r[i];
        r_sum += r[i];
    }

    if(m < l_sum || r_sum < m) {
        cout << -1 << endl;
        return;
    }

    // xを一つ増やすと、今まで入れた個数-cnt[x]だけ転倒数が増える
    //→xを3つ増やす場合は3*(今まで-cnt[x])。
    // {cnt[x], nxt, how_many, remain}

    int left = 0, right = INF;
    while(right-left > 1) {
        int mid = (left + right) / 2;
        int cur = 0;
        FOR(n) {
            cur += max(l[i], min(mid, r[i]));
        }
        // cout << mid << " " << cur << endl;
        if(cur >= m) right = mid;
        else left = mid;
    }

    // 後はright or right-1個まで使う数を増やしていく。
    int need = 0;
    FOR(n) need += max(l[i], min(right, r[i]));
    int no_need = need-m;
    int ans = 0;
    int cur_sum = 0;
    // for(int i = n-1; i >= 0; i--) {
    //     ans += l[i]*cur_sum;
    //     cur_sum += l[i];
    // }
    map<int, int> mp;
    FOR(n) {
        int use = max(l[i], min(right, r[i]));
        if(use == 0) continue;
        ++mp[0];
        if(use>l[i] && use == right && no_need > 0) {
            --use;
            --no_need;
        }
        --mp[use];
    }
    // for(auto e : mp) {
    //     cout << e.first << " " << e.second << endl;
    // }

    int num = 0, pre = -1;
    for(auto e : mp) {
        if(pre != -1 && num != 0) {
            ans += (cur_sum-pre)*((num-1)*(e.first-pre)+1) + ((num-1)*(e.first-pre))*((num-1)*(e.first-pre)+1)/2
                +(cur_sum-pre)*(e.first-pre-1)+(num-1)*(e.first-pre-1)*(e.first-pre)/2;
        }
        // cout << e.first << " " << ans << endl;

        cur_sum += num*(e.first-pre);
        num += e.second;
        pre = e.first;
    }

    cout << ans << endl;

    // cout << right << " " << need << " " << cur_sum << " " << no_need << endl;

    /*
    cur_sum-pre, cur_sum-pre+1, ... , cur_sum-pre+num-1, cur_sum-pre+num-1, cur_sum-pre+num, ... , cur_sum-pre+2*num-2
     cur_sum-pre, ... , cur_sum-pre+(num-1)*(e.first-pre)
    +cur_sum-pre+num-1+cur_sum-pre+2(num-1)+...+cur_sum-pre+(e.first-pre-1)

    = (cur_sum-pre)*((num-1)*(e.first-pre)+1) + ((num-1)*(e.first-pre))*((num-1)*(e.first-pre)+1)/2
     +(cur_sum-pre)*(e.first-pre-1)+(num-1)*(e.first-pre-1)*(e.first-pre)/2;
    */
}

signed main() {
    cin.tie(nullptr);
    ios::sync_with_stdio(false);
    int t;
    cin >> t;
    while(t--) solve();
    return 0;
}