#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 pb push_back
#define m0(x) memset(x,0,sizeof(x))
#define fill(x,y) memset(x,y,sizeof(x))
#define bg begin()
#define ed end()
#define all(x) x.bg,x.ed
//#define mp make_pair
#define vi vector<int>
#define vvi vector<vi>
#define vll vector<ll>
#define vvll vector<vll>
#define vs vector<string>
#define vvs vector<vs>
#define vc vector<char>
#define vvc vector<vc>
#define pii pair<int,int>
#define pllll pair<ll,ll>
#define vpii vector<pair<int,int>>
#define vpllll vector<pair<ll,ll>>
#define vpis vector<pair<int,string>>
#define vplls vector<pair<ll, string>>
#define vpsi vector<pair<string, int>>
#define vpsll vector<pair<string, ll>>

template<typename T>
void chmax(T &a, const T &b) {a = (a > b? a : b);}
template<typename T>
void chmin(T &a, const T &b) {a = (a < b? a : b);}

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

const ll INF = numeric_limits<long long>::max() / 2;
const ld pi = 3.1415926535897932384626433832795028;
const ll mod = 1e9 + 7;
int dx[] = {-1, 0, 1, 0, -1, -1, 1, 1};
int dy[] = {0, -1, 0, 1, -1, 1, -1, 1};

#define int long long

template <typename T>
struct RMQ {
    const T e = -INF;
    function<T(T, T)> fx = [](T x1, T x2) -> T { return max(x1, x2); };
    int n;
    vector<T> dat;
    RMQ(int n_) : n(), dat(n_ * 4, e) {
        int x = 1;
        while (n_ > x) {
            x *= 2;
        }
        n = x;
    }

    void add(int i, T x) {
        i += n - 1;
        dat[i] += x;
        while(i > 0) {
            i = (i - 1) / 2;
            dat[i] = fx(dat[i * 2 + 1], dat[i * 2 + 2]);
        }
    }
    void update(int i, T x) {
        i += n - 1;
        dat[i] = x;
        while (i > 0) {
            i = (i - 1) / 2;  // parent
            dat[i] = fx(dat[i * 2 + 1], dat[i * 2 + 2]);
        }
    }
    // the minimum element of [a,b)
    T query(int a, int b) { return query_sub(a, b, 0, 0, n); }
    T query_sub(int a, int b, int k, int l, int r) {
        if (r <= a || b <= l) {
            return e;
        } else if (a <= l && r <= b) {
            return dat[k];
        } else {
            T vl = query_sub(a, b, k * 2 + 1, l, (l + r) / 2);
            T vr = query_sub(a, b, k * 2 + 2, (l + r) / 2, r);
            return fx(vl, vr);
        }
    }
    int find_rightest(int a, int b, T x) { return find_rightest_sub(a, b, x, 0, 0, n); }
    int find_leftest(int a, int b, T x) { return find_leftest_sub(a, b, x, 0, 0, n); }
    int find_rightest_sub(int a, int b, T x, int k, int l, int r) {
        if (dat[k] > x || r <= a || b <= l) {  // 自分の値がxより大きい or [a,b)が[l,r)の範囲外ならreturn a-1
            return a - 1;
        } else if (k >= n - 1) {  // 自分が葉ならその位置をreturn
            return (k - (n - 1));
        } else {
            int vr = find_rightest_sub(a, b, x, 2 * k + 2, (l + r) / 2, r);
            if (vr != a - 1) {  // 右の部分木を見て a-1 以外ならreturn
                return vr;
            } else {  // 左の部分木を見て値をreturn
                return find_rightest_sub(a, b, x, 2 * k + 1, l, (l + r) / 2);
            }
        }
    }
    int find_leftest_sub(int a, int b, T x, int k, int l, int r) {
        if (dat[k] > x || r <= a || b <= l) {  // 自分の値がxより大きい or [a,b)が[l,r)の範囲外ならreturn b
            return b;
        } else if (k >= n - 1) {  // 自分が葉ならその位置をreturn
            return (k - (n - 1));
        } else {
            int vl = find_leftest_sub(a, b, x, 2 * k + 1, l, (l + r) / 2);
            if (vl != b) {  // 左の部分木を見て b 以外ならreturn
                return vl;
            } else {  // 右の部分木を見て値をreturn
                return find_leftest_sub(a, b, x, 2 * k + 2, (l + r) / 2, r);
            }
        }
    }

    inline T operator[](int i) {
        return query(i, i + 1);
    }

    void print() {
        for(int i = 0; i < n; i++) {
            cout << (*this)[i];
            if(i != n - 1) cout << ",";
        }
        cout << "\n";
    }
};

void solve() {
    // a[i] + b[i] < a[j] + b[j]

    // a[i] + b[j] < a[j] + b[i]
    // a[i] - b[i] < a[j] - b[j];

    int n;
    cin >> n;
    vi a(n), b(n);
    FOR(n) cin >> a[i];
    FOR(n) cin >> b[i];

    RMQ<int> tree1(n);
    RMQ<int> tree2(n);
    FOR(n) tree1.update(i, a[i] + b[i]);
    FOR(n) tree2.update(i, a[i] - b[i]);

    int ans = 0;
    int j = 0;

    for(int i = 0; i < n - 1; i++) {
        while(j < n && tree1.query(i, j) < tree1[j] && tree2.query(i, j) < tree2[j]) {
            j++;
        }
        ans += j - i - 1;
    }

    cout << ans << endl;
}

signed main() {
    cin.tie(nullptr);
    ios::sync_with_stdio(false);
    solve();
    return 0;
}