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#include <vector>
#include <iostream>
#include <sstream>
#include <math.h>
#include <sys/time.h>
#include <cstdlib>
#include <algorithm>
#include <cassert>
#include <cstring>
#include <fstream>
#include <set>
#include <map>
#include <iomanip>
#include <queue>

#define FOR(i,a,b)  for(__typeof(b) i=(a);i<(b);++i)
#define REP(i,a)    FOR(i,0,a)
#define FOREACH(x,c)   for(__typeof(c.begin()) x=c.begin();x != c.end(); x++)
#define ALL(c)      c.begin(),c.end()
#define CLEAR(c)    memset(c,0,sizeof(c))
#define SIZE(c) (int) ((c).size())

#define PB          push_back
#define MP          make_pair
#define X           first
#define Y           second

#define ULL         unsigned long long
#define LL          long long
#define LD          long double
#define II         pair<int, int>
#define DD         pair<double, double>
#define FF          pair<float,float>

#define VC	    vector
#define VI          VC<int>
#define VVI         VC<VI>
#define VVVI        VC<VVI>
#define VD          VC<double>
#define VS          VC<string>
#define VII         VC<II>
#define VDD         VC<DD>
#define VFF         VC<FF>
#define VVD         VC<VD>
#define VVVD        VC<VVD>
#define VULL         VC<ULL>

#define DB(a)       cerr << #a << ": " << a << endl;

using namespace std;

template<class T> void print(VC < T > v) {cerr << "[";if (SIZE(v) != 0) cerr << v[0]; FOR(i, 1, SIZE(v)) cerr << "," << v[i]; cerr << "]\n";}
template<class T> string i2s(T &x) { ostringstream o; o << x; return o.str(); }
VS split(string &s, char c = ' ') {VS all; int p = 0, np; while (np = s.find(c, p), np >= 0) {all.PB(s.substr(p, np - p)); p = np + 1;} if (p < SIZE(s)) all.PB(s.substr(p)); return all;}

int main(int argc, char *argv[]){
    int n;
    cin >> n;
    VI T(n);
    REP(i,n) cin >> T[i];
    int i = 0, j = n-1;
    LL cnt = 0;
    for(int med2 = n+1; med2 <= 2*n; med2++){
        while (T[i] < med2/2) i++;
        while (T[j] < med2/2) j--;
        // we re removing med2-n-1 elements
        int r = med2-n-1;
        // from two stacks of sizes i and n-1-j if possible
        if (i+n-1-j < r) continue;
        // no more than med2-n options no matter how large i and n-1-j are
        // and no more than i+1 or n-j 
        cnt += 1 + min(n-1-j,r) - max(0, r - i);
        //cout <<  med2 << " " << i << " " << j << " " << cnt << endl;
    }
    cout << n+1+n << " " << cnt << endl;
    return 0;
}