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, C[2000005];
int go[2000005];
vector<int> cnt[2000005];
vector<pair<int, int> > vecc[600005];
;
int main() {
  scanf("%d", &n);
  for (int i = 2; i <= 2000000; i++) {
    if (go[i]) continue;
    go[i] = i;
    for (int j = i + i; j <= 2000000; j += i) {
      go[j] = i;
    }
  }
  for (int i = 1; i <= n; i++) {
    int w;
    scanf("%d", &w);
    C[w]++;
  }
  int num = 0;
  bool ok = 0;
  for (int i = 2000000; i >= 2; i--) {
    if (C[i] == 0) continue;
    if (C[i] == 1) {
      if (cnt[i].empty()) {
        cnt[i].push_back(1);
        vecc[num++].push_back(make_pair(i, 1));
      } else {
        int cur = i - 1;
        vector<int> vec;
        while (cur > 1) {
          vec.push_back(go[cur]);
          cur /= go[cur];
        }
        sort(vec.begin(), vec.end());
        cur = -1;
        int ccnt = 0;
        for (int j = 0; j < vec.size(); j++) {
          if (j == 0) {
            cur = vec[j];
            ccnt = 1;
            continue;
          }
          if (cur != vec[j]) {
            cnt[cur].push_back(ccnt);
            vecc[num].push_back(make_pair(cur, ccnt));
            cur = vec[j];
            ccnt = 1;
          } else
            ccnt++;
        }
        if (cur > 1) {
          cnt[cur].push_back(ccnt);
          vecc[num++].push_back(make_pair(cur, ccnt));
        } else
          num++;
      }
    } else {
      if (C[i] > 2) ok = 1;
      cnt[i].push_back(1);
      vecc[num++].push_back(make_pair(i, 1));
      int cur = i - 1;
      vector<int> vec;
      while (cur > 1) {
        vec.push_back(go[cur]);
        cur /= go[cur];
      }
      sort(vec.begin(), vec.end());
      cur = -1;
      int ccnt = 0;
      for (int j = 0; j < vec.size(); j++) {
        if (j == 0) {
          cur = vec[j];
          ccnt = 1;
          continue;
        }
        if (cur != vec[j]) {
          cnt[cur].push_back(ccnt);
          vecc[num].push_back(make_pair(cur, ccnt));
          cur = vec[j];
          ccnt = 1;
        } else
          ccnt++;
      }
      if (cur > 1) {
        cnt[cur].push_back(ccnt);
        vecc[num++].push_back(make_pair(cur, ccnt));
      } else
        num++;
    }
  }
  for (int i = 2; i <= 2000000; i++) {
    sort(cnt[i].begin(), cnt[i].end());
    reverse(cnt[i].begin(), cnt[i].end());
  }
  for (int i = 0; i < num; i++) {
    for (int j = 0; j < vecc[i].size(); j++) {
      int A = vecc[i][j].first, B = vecc[i][j].second;
      if (cnt[A].size() == 1 || cnt[A][1] < B) goto fail;
    }
    ok = 1;
    break;
  fail:;
  }
  long long ans = 1;
  for (int i = 2; i <= 2000000; i++) {
    if (cnt[i].size())
      for (int j = 0; j < cnt[i][0]; j++) ans = ans * 1LL * i % 1000000007;
  }
  if (ok) ans = (ans + 1LL) % 1000000007;
  cout << ans << endl;
}