//模拟，突破口在木板高度是2000，记录每个木板高度出现的次数，以及能拼出来的高度的个数
#include <bits/stdc++.h>
using namespace std;

const int N = 2010, M = 4010;

int n, cnt[N], h[M];//cnt记录每个高度板子出现次数 h记录对应高度的板子的个数

int main() {
    scanf("%d", &n);
    for (int i = 0, x; i < n; ++ i) {
        scanf("%d", &x);
        cnt[x]++;
    }
    for (int i = 1; i <= 2000; ++ i)
        for (int j = i; j <= 2000; ++ j)
            if (i == j) h[i+j] += cnt[i]/2;
            else h[i+j] += min(cnt[i], cnt[j]);
    int len = 0, s = 0;
    for (int i = 1; i <= 4000; ++ i)
        if (h[i] > len) len = h[i], s = 1;
        else if (h[i] == len) s++;
    printf("%d %d", len, s);
    return 0;
}