//看看会不会爆int!数组会不会少了一维！//取物问题一定要小心先手胜利的条件#include <bits/stdc++.h>using namespace std;#define LL long long#define ALL(a) a.begin(), a.end()#define pb push_back#define mk make_pair#define fi first#define se secondconst int maxn = 5000 + 5;int tree[maxn], a[maxn];int big[maxn][maxn], big2[maxn][maxn];int n;int lowbit(int x) {return x & -x;}int sum(int x){    int ans = 0;    for (int i = x; i > 0; i -= lowbit(i)){        ans += tree[i];    }    return ans;}void add(int x, int val){    for (int i = x; i <= n; i += lowbit(i)){        tree[i] += val;    }}int main(){    scanf("%d", &n);    for (int i = 1; i <= n; i++) {        int u; scanf("%d", &u); u++;        a[i] = u;    }    int tot = 0;    for (int i = n; i >= 1; i--){        tot += sum(a[i]);        add(a[i], 1);    }    ///l->r的区间///区间内比他大的    for (int i = n; i > 0; i--){        memset(tree, 0, sizeof(tree));        for (int j = n; j >= i;j--){///都取不到边界            if (a[j] > a[i]) add(a[j], 1);            big[i][j] = sum(n) - sum(a[i] - 1);        }    }    ///r->l的区间///区间内比他大的    for (int i = 1; i <= n; i++){        memset(tree, 0, sizeof(tree));        for (int j = 1; j <= i; j++){            if (a[j] > a[i]) add(a[j], 1);            big2[i][j] = sum(n) - sum(a[i] - 1);        }    }    int mintot = tot;    int cnt = 0;    for (int i = 1; i <= n; i++){///left        for (int j = i + 1;  j <= n; j++){///right            if (a[i] < a[j]) continue;            int t1 = 2 * (big[i][i + 1] - big[i][j]) - (j - (i + 1));            int t2 = j - (i + 1) - 2 * (big2[j][j - 1] - big2[j][i]);            int tmp = tot + t1 + t2 - 1;            if (tmp < mintot) cnt = 1, mintot = tmp;            else if (tmp == mintot) cnt++;        }    }    printf("%d %d\n", mintot, cnt);    return 0;}