//经典树状数组求逆序对数量
//求出 左边比当前数大的数的和，求出右边比当前数大的和，乘法原理就是V，^同理

#include<bits/stdc++.h>

#define lowbit(x) x & (-x)

using namespace std;
typedef long long ll;
const int N = 2e5 + 10, inf = 0x3f3f3f3f;

int n;
int a[N], tr[N];
int bigger[N], lower[N];

void add(int id, int x)
{
    for(int i = id; i <= n; i += lowbit(i)) tr[i] += x;
}

ll qurry(int l, int r)//注意，查询区间是（l，r]
{
    int tolr = 0, toll = 0;
    for(int i = r; i; i -= lowbit(i)) tolr += tr[i];
    for(int i = l; i; i -= lowbit(i)) toll += tr[i];
    return tolr - toll;
}

int main()
{
    ios::sync_with_stdio(false);
    cin.tie(nullptr), cout.tie(nullptr);

    cin >> n;
    for(int i = 1; i <= n; i ++) cin >> a[i];

    for(int i = 1; i <= n; i ++)//左往右
    {
        int y = a[i];
        bigger[i] = qurry(y, n);//比当前数大的数字的个数，就是树状数组中[y + 1, n]中的数字和
        lower[i] = qurry(0, y - 1);//比当前数字小的数的个数，就是树状数组中[1, y - 1]中的数字和
        add(y, 1);
    }

    memset(tr, 0, sizeof tr);
    ll res1 = 0, res2 = 0;
    for(int i = n; i; i --)//右往左
    {
        int y = a[i];
        res1 += bigger[i] * (ll)qurry(y, n);
        res2 += lower[i] * (ll)qurry(0, y - 1);
        add(y, 1);
    }
    cout << res1 << ' ' << res2 << '\n';
    return 0;
}