#include <bits/stdc++.h>
using namespace std;
pair<int, int> h[100010], h1[100010];
long long cnt[100010], pre[100010], sum[100010];
int num[100010], mx[100010], n, tt, t1;
inline int rd() {
  int x = 0;
  char ch = getchar();
  for (; ch < '0' || ch > '9'; ch = getchar())
    ;
  for (; ch >= '0' && ch <= '9'; ch = getchar()) x = x * 10 + ch - '0';
  return x;
}
inline int gcd(int x, int y) { return (!y) ? x : gcd(y, x % y); }
inline long long calc(long long l, long long r) {
  return (l + r) * (r - l + 1) / 2;
}
inline long long anol_gcd(long long n, long long a, long long b, long long c) {
  if (n < 0) return 0;
  if (!a) return (b / c) * (n + 1);
  if (a >= c || b >= c)
    return (a / c) * n * (n + 1) / 2 + (b / c) * (n + 1) +
           anol_gcd(n, a % c, b % c, c);
  long long hh = (a * n + b) / c;
  return n * hh - anol_gcd(hh - 1, c, c - b - 1, a);
}
inline long long gao(int x, int y, long long mid) {
  if (!cnt[x] || !cnt[y]) return 0;
  mid -= pre[y - 1] - pre[x];
  if (mid < 0) return 0;
  long long mn = (mid - cnt[y] * y) / x;
  mn = max(mn, 0LL);
  if (mn >= cnt[x]) return cnt[x] * cnt[y];
  long long res = mn * cnt[y];
  mn++;
  long long mx = mid / x;
  mx = min(mx, cnt[x]);
  if (mn <= mx) res += anol_gcd(mx - mn, x, mid - mx * x, y);
  return res;
}
inline long long check(long long mid) {
  long long res = 0;
  int now = 1;
  for (int i = 1; i <= 100000; i++) {
    now = max(now, i);
    long long hh = min(mid / i, cnt[i]);
    if (cnt[i]) res += calc(cnt[i] - hh + 1, cnt[i]);
    while (now + 1 <= 100000 && pre[now + 1] - pre[i - 1] <= mid) now++;
    res += (sum[now] - sum[i]) * cnt[i];
    mx[i] = now;
  }
  for (int i = 1; i <= 100000; i++)
    for (int j = mx[i] + 1; j <= min(mx[i + 1] + 1, 100000); j++)
      res += gao(i, j, mid);
  return res;
}
int main() {
  n = rd();
  for (int i = 1; i <= n; i++) num[i] = rd();
  cnt[num[1]]++;
  h[tt = 1] = pair<int, int>(1, num[1]);
  for (int i = 2; i <= n; i++) {
    t1 = tt;
    for (int j = 1; j <= t1; j++)
      h1[j] = h[j], h1[j].second = gcd(h1[j].second, num[i]);
    h1[++t1] = pair<int, int>(1, num[i]);
    tt = 0;
    for (int j = 1; j <= t1;) {
      int k = j, now = 0;
      for (; k <= t1 && h1[k].second == h1[j].second; k++) now += h1[k].first;
      h[++tt] = pair<int, int>(now, h1[j].second);
      j = k;
    }
    for (int j = 1; j <= tt; j++) cnt[h[j].second] += h[j].first;
  }
  for (int i = 1; i <= 100000; i++)
    pre[i] = pre[i - 1] + cnt[i] * i, sum[i] = sum[i - 1] + cnt[i];
  long long all = (long long)n * (n + 1) / 2;
  all = all * (all + 1) / 2;
  all = (all + 1) / 2;
  long long l = 0, r = (long long)n * (n + 1) / 2 * 100000;
  while (l < r) {
    long long mid = (l + r) >> 1;
    if (check(mid) >= all)
      r = mid;
    else
      l = mid + 1;
  }
  printf("%lld\n", l);
  return 0;
}
