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>
int mxpower[2000001], cnt[2000001], n, a[2000001], now, P = 1000000007;
bool choose[2000001];
bool check(int x) {
  for (int i = 2; i * i <= x; i++) {
    int tem = 0;
    while (x % i == 0) x /= i, ++tem;
    if (mxpower[i] == tem && cnt[i] == 1) return 0;
  }
  if (x > 1)
    if (mxpower[x] == 1 && cnt[x] == 1) return 0;
  return 1;
}
int qsm(int a, int b) {
  int ans = 1;
  while (b) {
    if (b & 1) ans = 1ll * ans * a % P;
    a = 1ll * a * a % P;
    b >>= 1;
  }
  return ans;
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", a + i);
  std::sort(a + 1, a + n + 1);
  for (int i = n; i; i--)
    if (mxpower[a[i]]) {
      choose[i] = 1;
      int tem = a[i] - 1;
      for (int j = 2; j * j <= tem; j++)
        if (tem % j == 0) {
          int po = 0;
          while (tem % j == 0) tem /= j, ++po;
          if (mxpower[j] == po)
            ++cnt[j];
          else if (mxpower[j] < po)
            mxpower[j] = po, cnt[j] = 1;
        }
      if (tem > 1)
        if (mxpower[tem] == 1)
          ++cnt[tem];
        else if (mxpower[tem] < 1)
          mxpower[tem] = cnt[tem] = 1;
    } else
      mxpower[a[i]] = cnt[a[i]] = 1;
  for (int i = n; i; i--)
    if (choose[i]) {
      if (check(a[i] - 1) && !now) {
        now = 1;
        int tem = a[i] - 1;
        for (int j = 2; j * j <= tem; j++)
          if (tem % j == 0) {
            int po = 0;
            while (tem % j == 0) tem /= j, ++po;
            if (mxpower[j] == po) --cnt[j];
          }
        if (tem > 1)
          if (mxpower[tem] == 1) --cnt[tem];
      }
    } else {
      if (cnt[a[i]] > 1) --cnt[a[i]], now = 1;
    }
  int ans = 1;
  for (int i = 2; i <= 2000000; i++) ans = 1ll * ans * qsm(i, mxpower[i]) % P;
  printf("%d", (ans + now) % P);
}