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 maxm = 2e6 + 10;
const int M = 2e6 + 1;
const int mod = 1e9 + 7;
const int maxn = 2e5 + 10;
int vis[maxm], pri[maxm / 5], psz, nxt[maxm];
int a[maxn], n, flg = 0, b[maxn];
map<int, pair<int, int> > st;
int qpow(int a, int b) {
  int ans = 1;
  for (; b; b >>= 1, a = 1ll * a * a % mod)
    if (b & 1) ans = 1ll * ans * a % mod;
  return ans;
}
void ins(int x) {
  while (x > 1) {
    int y = nxt[x], z = 0;
    while (x % y == 0) x /= y, z++;
    if (z > st[y].first)
      st[y] = pair<int, int>(z, 1);
    else if (z == st[y].first)
      st[y].second++;
  }
}
int chk(int x) {
  if (x == 1) return 1;
  int ret = 0;
  int c = 1;
  while (x > 1) {
    int y = nxt[x], z = 0;
    while (x % y == 0) x /= y, z++;
    if (!((st[y].second > 1 && st[y].first == z) || (st[y].first > z))) c = 0;
  }
  return c;
}
int main() {
  for (int i = 2; i <= M; ++i) {
    if (!vis[i]) {
      pri[++psz] = i;
      nxt[i] = i;
    }
    for (int j = 1; j <= psz && pri[j] * i <= M; ++j) {
      vis[i * pri[j]] = 1;
      nxt[i * pri[j]] = pri[j];
      if (i % pri[j] == 0) break;
    }
  }
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) scanf("%d", &a[i]);
  sort(a + 1, a + n + 1, greater<int>());
  for (int i = 1; i <= n; ++i)
    if (a[i] > 1) {
      if (st[a[i]].first >= 1) {
        ins(a[i] - 1), b[i] = 0;
      } else
        ins(a[i]), b[i] = 1;
    }
  int ans = 1;
  for (auto c : st) ans = 1ll * ans * qpow(c.first, c.second.first) % mod;
  for (int i = 1; i <= n; ++i)
    if (a[i] > 1) {
      if (b[i] == 0) {
        flg |= chk(a[i] - 1);
      } else
        flg |= chk(a[i]);
    }
  printf("%d", (ans + flg) % mod);
}