#include <bits/stdc++.h>
int F() {
  register int a;
  register char c;
  while (c = getchar(), c < '0' || c > '9')
    ;
  for (a = c - '0'; c = getchar(), c >= '0' && c <= '9'; a = a * 10 + c - '0')
    ;
  return a;
}
const int N = 2e5, M = 2e6 + 5, P = 1e9 + 7;
int p[N], q[N], mp[M], pr[M], times[M], maxcnt[M];
void clcm(register int x) {
  register int id, cur, cnt;
  while (x != 1) {
    cur = pr[id = mp[x]];
    for (cnt = 0; x % cur == 0; x /= cur) ++cnt;
    if (times[id] == cnt) ++maxcnt[id];
    if (((cnt) > times[id] ? times[id] = (cnt), 1 : 0)) maxcnt[id] = 1;
  }
}
bool check(register int x) {
  register int id, cur, cnt;
  register bool ans = 1;
  while (x != 1) {
    cur = pr[id = mp[x]];
    for (cnt = 0; x % cur == 0; x /= cur) ++cnt;
    if (times[id] == cnt && maxcnt[id] == 1) ans = 0;
  }
  return ans;
}
int main() {
  register int n, m, i, j, cnt = 0, ans = 1;
  register bool inc = 0;
  n = F();
  for (i = 0; i < n; ++i) p[i] = F();
  std::sort(p, p + n, std::greater<int>());
  m = p[0];
  for (i = 2; i <= m; ++i) {
    if (!mp[i]) pr[mp[i] = ++cnt] = i;
    for (j = 1; j <= cnt && i * pr[j] <= m; ++j) {
      mp[i * pr[j]] = j;
      if (i % pr[j] == 0) break;
    }
  }
  for (i = 0; i < n; ++i) clcm(q[i] = p[i] - (bool)times[mp[p[i]]]);
  for (i = 1; i <= cnt; ++i) {
    for (j = 1; j <= times[i]; ++j) {
      ans = (long long)ans * pr[i] % P;
    }
  }
  for (i = 0; i < n; ++i) inc |= check(q[i]);
  printf("%d\n", (ans + inc) % P);
  return 0;
}

### Prompt

Please provide a Cpp coded solution to the problem described below:

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].

### Response

```cpp
#include <bits/stdc++.h>
int F() {
  register int a;
  register char c;
  while (c = getchar(), c < '0' || c > '9')
    ;
  for (a = c - '0'; c = getchar(), c >= '0' && c <= '9'; a = a * 10 + c - '0')
    ;
  return a;
}
const int N = 2e5, M = 2e6 + 5, P = 1e9 + 7;
int p[N], q[N], mp[M], pr[M], times[M], maxcnt[M];
void clcm(register int x) {
  register int id, cur, cnt;
  while (x != 1) {
    cur = pr[id = mp[x]];
    for (cnt = 0; x % cur == 0; x /= cur) ++cnt;
    if (times[id] == cnt) ++maxcnt[id];
    if (((cnt) > times[id] ? times[id] = (cnt), 1 : 0)) maxcnt[id] = 1;
  }
}
bool check(register int x) {
  register int id, cur, cnt;
  register bool ans = 1;
  while (x != 1) {
    cur = pr[id = mp[x]];
    for (cnt = 0; x % cur == 0; x /= cur) ++cnt;
    if (times[id] == cnt && maxcnt[id] == 1) ans = 0;
  }
  return ans;
}
int main() {
  register int n, m, i, j, cnt = 0, ans = 1;
  register bool inc = 0;
  n = F();
  for (i = 0; i < n; ++i) p[i] = F();
  std::sort(p, p + n, std::greater<int>());
  m = p[0];
  for (i = 2; i <= m; ++i) {
    if (!mp[i]) pr[mp[i] = ++cnt] = i;
    for (j = 1; j <= cnt && i * pr[j] <= m; ++j) {
      mp[i * pr[j]] = j;
      if (i % pr[j] == 0) break;
    }
  }
  for (i = 0; i < n; ++i) clcm(q[i] = p[i] - (bool)times[mp[p[i]]]);
  for (i = 1; i <= cnt; ++i) {
    for (j = 1; j <= times[i]; ++j) {
      ans = (long long)ans * pr[i] % P;
    }
  }
  for (i = 0; i < n; ++i) inc |= check(q[i]);
  printf("%d\n", (ans + inc) % P);
  return 0;
}

```