Problem: You are given a tuple generator f^{(k)} = (f_1^{(k)}, f_2^{(k)}, ..., f_n^{(k)}), where f_i^{(k)} = (a_i ⋅ f_i^{(k - 1)} + b_i) mod p_i and f^{(0)} = (x_1, x_2, ..., x_n). Here x mod y denotes the remainder of x when divided by y. All p_i are primes.

One can see that with fixed sequences x_i, y_i, a_i the tuples f^{(k)} starting from some index will repeat tuples with smaller indices. Calculate the maximum number of different tuples (from all f^{(k)} for k ≥ 0) that can be produced by this generator, if x_i, a_i, b_i are integers in the range [0, p_i - 1] and can be chosen arbitrary. The answer can be large, so print the remainder it gives when divided by 10^9 + 7

Input

The first line contains one integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the number of elements in the tuple.

The second line contains n space separated prime numbers — the modules p_1, p_2, …, p_n (2 ≤ p_i ≤ 2 ⋅ 10^6).

Output

Print one integer — the maximum number of different tuples modulo 10^9 + 7.

Examples

Input

4
2 3 5 7


Output

210


Input

3
5 3 3


Output

30

Note

In the first example we can choose next parameters: a = [1, 1, 1, 1], b = [1, 1, 1, 1], x = [0, 0, 0, 0], then f_i^{(k)} = k mod p_i.

In the second example we can choose next parameters: a = [1, 1, 2], b = [1, 1, 0], x = [0, 0, 1].

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int inf = 1e9;
const long long Inf = 1e18;
const int N = 2e6 + 10;
const int mod = 1e9 + 7;
int gi() {
  int x = 0, o = 1;
  char ch = getchar();
  while ((ch < '0' || ch > '9') && ch != '-') ch = getchar();
  if (ch == '-') o = -1, ch = getchar();
  while (ch >= '0' && ch <= '9') x = x * 10 + ch - '0', ch = getchar();
  return x * o;
}
template <typename T>
bool chkmax(T &a, T b) {
  return a < b ? a = b, 1 : 0;
};
template <typename T>
bool chkmin(T &a, T b) {
  return a > b ? a = b, 1 : 0;
};
int add(int a, int b) { return a + b >= mod ? a + b - mod : a + b; }
int sub(int a, int b) { return a - b < 0 ? a - b + mod : a - b; }
void inc(int &a, int b) { a = (a + b >= mod ? a + b - mod : a + b); }
void dec(int &a, int b) { a = (a - b < 0 ? a - b + mod : a - b); }
int n, p[N], a[N], pw[N], cnt[N];
int pri[N], tot = 0, d[N];
bool vis[N];
void init() {
  const int n = 2e6;
  for (int i = 2; i <= n; i++) {
    if (!vis[i]) pri[++tot] = i, d[i] = i;
    for (int j = 1; j <= tot && i * pri[j] <= n; j++) {
      vis[i * pri[j]] = 1;
      d[i * pri[j]] = pri[j];
      if (i % pri[j] == 0) break;
    }
  }
}
int qpow(int a, int b) {
  int ret = 1;
  while (b) {
    if (b & 1) ret = 1ll * ret * a % mod;
    a = 1ll * a * a % mod;
    b >>= 1;
  }
  return ret;
}
int main() {
  init();
  cin >> n;
  for (int i = 1; i <= n; i++) p[i] = gi();
  sort(p + 1, p + n + 1);
  for (int i = n; i; i--) {
    if (!pw[p[i]])
      pw[p[i]] = cnt[p[i]] = 1, a[i] = p[i];
    else {
      int x = p[i] - 1;
      a[i] = p[i] - 1;
      while (x > 1) {
        int y = d[x], sum = 0;
        while (x % y == 0) x /= y, sum++;
        if (pw[y] < sum)
          pw[y] = sum, cnt[y] = 1;
        else if (pw[y] == sum)
          cnt[y]++;
      }
    }
  }
  int tag = 0;
  for (int i = n; i; i--) {
    int x = a[i], fl = 1;
    while (x > 1) {
      int y = d[x], sum = 0;
      while (x % y == 0) x /= y, sum++;
      if (sum == pw[y] && cnt[y] == 1) fl = 0;
    }
    tag |= fl;
  }
  int ans = 1;
  for (int i = 2; i <= 2000000; i++) ans = 1ll * ans * qpow(i, pw[i]) % mod;
  printf("%d\n", add(ans, tag));
  return 0;
}