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 mod = 1000000007;
int n, p[200001], a[2000001], c[2000001][10], cn[2000001], d[200001],
    num[2000001];
bool bo[2000001], BO;
int quickmi(int a, int b) {
  int t = 1;
  while (b) {
    if (b & 1) t = 1ll * t * a % mod;
    a = 1ll * a * a % mod;
    b >>= 1;
  }
  return t;
}
int main() {
  memset(bo, 1, sizeof(bo));
  memset(cn, 0, sizeof(cn));
  for (int i = 2; i <= 2000000; i++)
    if (bo[i]) {
      c[i][++cn[i]] = i;
      for (int j = (i << 1); j <= 2000000; j += i) bo[j] = 0, c[j][++cn[j]] = i;
    }
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", &p[i]);
  sort(p + 1, p + n + 1);
  int ans = 1;
  BO = 0;
  memset(a, 0, sizeof(a));
  memset(num, 0, sizeof(num));
  for (int i = n; i; i--) {
    int y;
    bool boo = 0;
    if ((i == 1 || p[i] != p[i - 1]) && (!a[p[i]]))
      y = p[i];
    else
      y = p[i] - 1;
    for (int j = 1; j <= cn[y]; j++) {
      int x = y, cnt = 0;
      while (x % c[y][j] == 0) x /= c[y][j], cnt++;
      if (cnt > a[c[y][j]])
        a[c[y][j]] = cnt, boo = 1, num[c[y][j]] = 1;
      else if (cnt == a[c[y][j]])
        num[c[y][j]]++;
    }
    d[i] = y;
  }
  for (int i = 1; i <= n; i++) {
    bool boo = 0;
    int y = d[i];
    for (int j = 1; j <= cn[y]; j++) {
      int x = y, cnt = 0;
      while (x % c[y][j] == 0) x /= c[y][j], cnt++;
      if (cnt == a[c[y][j]] && num[c[y][j]] == 1) boo = 1;
    }
    if (!boo) {
      BO = 1;
      break;
    }
  }
  for (int i = 2; i <= 2000000; i++) ans = 1ll * ans * quickmi(i, a[i]) % mod;
  if (BO) ans = (ans + 1) % mod;
  printf("%d\n", ans);
  return 0;
}