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;
int n, p[666666], mr[2000555];
bool np[2000555];
int fy[2000555];
int ny[6000000], yy[6000000], an, cm[2000555];
const int MOD = 1e9 + 7;
map<int, int> sb;
bool go(int g) {
  int su = 0;
  for (int t = fy[g]; t; t = ny[t]) {
    int y = 0;
    while (g % yy[t] == 0) ++y, g /= yy[t];
    if (mr[yy[t]] >= y) continue;
    su = 1;
    mr[yy[t]] = y;
  }
  return su;
}
int mc[2000555];
void AA(int g) {
  for (int t = fy[g]; t; t = ny[t]) {
    int y = 0;
    while (g % yy[t] == 0) ++y, g /= yy[t];
    if (mr[yy[t]] == y) ++mc[yy[t]];
  }
}
bool BB(int g) {
  bool must = 0;
  for (int t = fy[g]; t; t = ny[t]) {
    int y = 0;
    while (g % yy[t] == 0) ++y, g /= yy[t];
    if (mr[yy[t]] == y) must |= mc[yy[t]] == 1;
  }
  return must;
}
long long zy[666666];
int zn = 0;
int main() {
  for (int i = 2; i <= 2000000; ++i)
    if (!np[i])
      for (int j = i; j <= 2000000; j += i)
        ++an, yy[an] = i, ny[an] = fy[j], fy[j] = an, np[j] = 1;
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) scanf("%d", p + i), ++sb[-p[i]];
  int aa = 0;
  for (auto g : sb) {
    if (g.second < 2) continue;
    int w = -g.first;
    if (!go(w))
      aa = 1;
    else
      zy[++zn] = w;
    for (int j = 1; j < g.second; ++j)
      if (!go(w - 1))
        aa = 1;
      else
        zy[++zn] = w - 1;
  }
  for (auto g : sb) {
    if (g.second != 1) continue;
    int w = -g.first;
    if (go(w))
      zy[++zn] = w;
    else if (go(w - 1))
      zy[++zn] = w - 1;
    else
      aa = 1;
  }
  for (int i = 1; i <= zn; ++i) AA(zy[i]);
  for (int i = 1; i <= zn; ++i)
    if (!BB(zy[i])) aa = 1;
  long long ans = 1;
  for (int i = 1; i <= 2000000; ++i)
    while (mr[i]--) ans = ans * i % MOD;
  ans += aa;
  ans = (ans % MOD + MOD) % MOD;
  cout << ans << "\n";
}