#include <bits/stdc++.h>
using pii = std::pair<int, int>;
using ll = long long;
const int lim = 2e6 + 22;
const int maxn = 2e6 + 555;
const int mod = 1e9 + 7;
int n, D[maxn], pw[maxn], cnt[maxn], p[maxn];
std::vector<pii> v[maxn];
int qpow(int a, int b) {
  if (b == 0) return 1;
  ll d = qpow(a, b >> 1);
  d = d * d % mod;
  if (b & 1) d = d * a % mod;
  return d;
}
void inc(int x, int y) {
  if (pw[x] < y) {
    pw[x] = y;
    cnt[x] = 1;
  } else if (pw[x] == y)
    cnt[x]++;
}
int useful(int x, int y) {
  if (y < pw[x]) return 0;
  if (y == pw[x] && cnt[x] > 1) return 0;
  return 1;
}
std::vector<pii> work(int x) {
  std::vector<pii> v;
  while (x != 1) {
    int d = D[x];
    int c = 0;
    while (x % d == 0) {
      x /= d;
      c++;
    }
    v.push_back({d, c});
  }
  return v;
}
int main() {
  scanf("%d", &n);
  for (int i = 2; i <= lim; ++i) {
    if (D[i]) continue;
    for (int j = i; j <= lim; j += i) D[j] = i;
  }
  for (int i = 1; i <= n; ++i) scanf("%d", &p[i]);
  std::sort(p + 1, p + n + 1, [](int a, int b) { return a > b; });
  for (int i = 1; i <= n; ++i) {
    if (!pw[p[i]]) {
      inc(p[i], 1);
      v[i] = work(p[i]);
    } else {
      v[i] = work(p[i] - 1);
      for (auto p : v[i]) {
        int w = p.first, c = p.second;
        inc(w, c);
      }
    }
  }
  int ans = 1;
  for (int i = 1; i <= lim; ++i) {
    if (pw[i]) ans = (ll)ans * qpow(i, pw[i]) % mod;
  }
  for (int i = 1; i <= n; ++i) {
    int flag = 0;
    for (auto p : v[i])
      if (useful(p.first, p.second)) {
        flag = 1;
        break;
      }
    if (!flag) {
      ans = (ans + 1) % mod;
      break;
    }
  }
  printf("%d", ans);
  return 0;
}

### Prompt

Please formulate a CPP solution to the following 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].

### Response

```cpp
#include <bits/stdc++.h>
using pii = std::pair<int, int>;
using ll = long long;
const int lim = 2e6 + 22;
const int maxn = 2e6 + 555;
const int mod = 1e9 + 7;
int n, D[maxn], pw[maxn], cnt[maxn], p[maxn];
std::vector<pii> v[maxn];
int qpow(int a, int b) {
  if (b == 0) return 1;
  ll d = qpow(a, b >> 1);
  d = d * d % mod;
  if (b & 1) d = d * a % mod;
  return d;
}
void inc(int x, int y) {
  if (pw[x] < y) {
    pw[x] = y;
    cnt[x] = 1;
  } else if (pw[x] == y)
    cnt[x]++;
}
int useful(int x, int y) {
  if (y < pw[x]) return 0;
  if (y == pw[x] && cnt[x] > 1) return 0;
  return 1;
}
std::vector<pii> work(int x) {
  std::vector<pii> v;
  while (x != 1) {
    int d = D[x];
    int c = 0;
    while (x % d == 0) {
      x /= d;
      c++;
    }
    v.push_back({d, c});
  }
  return v;
}
int main() {
  scanf("%d", &n);
  for (int i = 2; i <= lim; ++i) {
    if (D[i]) continue;
    for (int j = i; j <= lim; j += i) D[j] = i;
  }
  for (int i = 1; i <= n; ++i) scanf("%d", &p[i]);
  std::sort(p + 1, p + n + 1, [](int a, int b) { return a > b; });
  for (int i = 1; i <= n; ++i) {
    if (!pw[p[i]]) {
      inc(p[i], 1);
      v[i] = work(p[i]);
    } else {
      v[i] = work(p[i] - 1);
      for (auto p : v[i]) {
        int w = p.first, c = p.second;
        inc(w, c);
      }
    }
  }
  int ans = 1;
  for (int i = 1; i <= lim; ++i) {
    if (pw[i]) ans = (ll)ans * qpow(i, pw[i]) % mod;
  }
  for (int i = 1; i <= n; ++i) {
    int flag = 0;
    for (auto p : v[i])
      if (useful(p.first, p.second)) {
        flag = 1;
        break;
      }
    if (!flag) {
      ans = (ans + 1) % mod;
      break;
    }
  }
  printf("%d", ans);
  return 0;
}

```