Problem: In a far away galaxy there is war again. The treacherous Republic made k precision strikes of power ai on the Empire possessions. To cope with the republican threat, the Supreme Council decided to deal a decisive blow to the enemy forces. 

To successfully complete the conflict, the confrontation balance after the blow should be a positive integer. The balance of confrontation is a number that looks like <image>, where p = n! (n is the power of the Imperial strike), <image>. After many years of war the Empire's resources are low. So to reduce the costs, n should be a minimum positive integer that is approved by the commanders.

Help the Empire, find the minimum positive integer n, where the described fraction is a positive integer.

Input

The first line contains integer k (1 ≤ k ≤ 106). The second line contains k integers a1, a2, ..., ak (1 ≤ ai ≤ 107).

Output

Print the minimum positive integer n, needed for the Empire to win.

Please, do not use the %lld to read or write 64-but integers in С++. It is preferred to use the cin, cout streams or the %I64d specificator.

Examples

Input

2
1000 1000


Output

2000

Input

1
2


Output

2

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const double PI = acos(-1);
const double eps = 1e-12;
const int inf = 2000000000;
const long long int infLL = (long long int)2e18;
long long int MOD = 1000000007;
int MOD1 = 1000000007;
int MOD2 = 1000000009;
inline bool checkBit(long long int n, long long int i) {
  return n & (1LL << i);
}
inline long long int setBit(long long int n, long long int i) {
  return n | (1LL << i);
  ;
}
inline long long int resetBit(long long int n, long long int i) {
  return n & (~(1LL << i));
}
int dx[] = {0, 0, +1, -1};
int dy[] = {+1, -1, 0, 0};
inline bool EQ(double a, double b) { return fabs(a - b) < 1e-9; }
inline bool isLeapYear(long long int year) {
  return (year % 400 == 0) || (year % 4 == 0 && year % 100 != 0);
}
inline void normal(long long int &a) {
  a %= MOD;
  (a < 0) && (a += MOD);
}
inline long long int modMul(long long int a, long long int b) {
  a %= MOD, b %= MOD;
  normal(a), normal(b);
  return (a * b) % MOD;
}
inline long long int modAdd(long long int a, long long int b) {
  a %= MOD, b %= MOD;
  normal(a), normal(b);
  return (a + b) % MOD;
}
inline long long int modSub(long long int a, long long int b) {
  a %= MOD, b %= MOD;
  normal(a), normal(b);
  a -= b;
  normal(a);
  return a;
}
inline long long int modPow(long long int b, long long int p) {
  long long int r = 1LL;
  while (p) {
    if (p & 1) r = modMul(r, b);
    b = modMul(b, b);
    p >>= 1LL;
  }
  return r;
}
inline long long int modDiv(long long int a, long long int b) {
  return modMul(a, modPow(b, MOD - 2));
}
bool comp(const pair<long long int, pair<long long int, long long int> > &p1,
          const pair<long long int, pair<long long int, long long int> > &p2) {
  return p1.first > p2.first;
}
bool comp1(const pair<long long int, long long int> &p1,
           const pair<long long int, long long int> &p2) {
  if (p1.first == p2.first) {
    return p1.second > p2.second;
  }
  return p1.first < p2.first;
}
long long int converter(string a) {
  long long int i, mul = 1, r, t, ans = 0LL;
  if (a.length() == 0) return 0;
  for (i = a.length() - 1; i >= 0; i--) {
    t = a[i] - '0';
    r = t % 10;
    ans += (mul * r);
    mul = mul * 10;
  }
  return ans;
}
int msb(unsigned x) {
  union {
    double a;
    int b[2];
  };
  a = x;
  return (b[1] >> 20) - 1023;
}
const int MAX = 10000005;
long long int cnt[MAX + 5], n, p[MAX + 5];
vector<long long int> prime;
void sieve() {
  long long int i, j, x = sqrt(MAX) + 1;
  for (i = 2; i < MAX; ++i) p[i] = i;
  for (i = 2; i <= x; ++i)
    if (p[i] == i) {
      for (j = i * i; j < MAX; j += i) p[j] = min(p[j], i);
    }
  for (i = 2; i < MAX; ++i)
    if (p[i] == i) prime.push_back(i);
}
bool oka(long long int now) {
  for (int i = 0; i < prime.size(); ++i) {
    long long int x = now, cur = 0;
    if (cnt[prime[i]] != 0) {
      while (1) {
        if (x < prime[i]) break;
        cur += (x / prime[i]);
        x = x / prime[i];
      }
      if (cur < cnt[prime[i]]) return false;
    }
  }
  return true;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  long long int i, x;
  sieve();
  cin >> n;
  for (i = 1; i <= n; ++i) {
    cin >> x;
    cnt[x]++;
  }
  for (i = MAX - 1; i >= 2; --i) cnt[i] += cnt[i + 1];
  for (i = MAX - 1; i >= 2; --i)
    if (p[i] != i) {
      cnt[p[i]] += cnt[i];
      cnt[i / p[i]] += cnt[i];
    }
  long long int low = 1, high = (long long int)2e13, mid, ans = infLL;
  while (low <= high) {
    mid = (low + high) >> 1LL;
    if (oka(mid) == true) {
      ans = min(ans, mid);
      high = mid - 1;
    } else
      low = mid + 1;
  }
  cout << ans << '\n';
  return 0;
}