Let us denote by d(n) the sum of all divisors of the number n, i.e. d(n) = ∑_{k | n} k.

For example, d(1) = 1, d(4) = 1+2+4=7, d(6) = 1+2+3+6=12.

For a given number c, find the minimum n such that d(n) = c.

Input

The first line contains one integer t (1 ≤ t ≤ 10^4). Then t test cases follow.

Each test case is characterized by one integer c (1 ≤ c ≤ 10^7).

Output

For each test case, output: 

  * "-1" if there is no such n that d(n) = c; 
  * n, otherwise. 

Example

Input


12
1
2
3
4
5
6
7
8
9
10
39
691


Output


1
-1
2
3
-1
5
4
7
-1
-1
18
-1
#include <bits/stdc++.h>
using namespace std;
int64_t _sieve_size;
bitset<100010> bs;
vector<int64_t> primes;
void sieve(int64_t upperbound) {
  _sieve_size = upperbound + 1;
  bs.set();
  bs[0] = bs[1] = 0;
  for (int64_t i = 2; i <= _sieve_size; ++i) {
    if (bs[i]) {
      for (int64_t j = i * i; j <= _sieve_size; ++j) {
        bs[j] = 0;
      }
      primes.push_back(i);
    }
  }
  return;
}
int64_t sumDiv(int64_t N) {
  int64_t PF_idx = 0, PF = primes[PF_idx], ans = 1;
  while (PF * PF <= N) {
    int64_t power = 0;
    while (N % PF == 0) {
      N /= PF;
      ++power;
    }
    ans *= (pow(PF, power + 1) - 1) / (PF - 1);
    PF = primes[++PF_idx];
  }
  if (N != 1) ans *= (pow(N, 2) - 1) / (N - 1);
  return ans;
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  sieve(100000);
  vector<int64_t> S(10000010, -1);
  for (int64_t i = 1; i <= 10000000; ++i) {
    int64_t sum = sumDiv(i);
    if (sum > 10000000) continue;
    if (S[sum] == -1) {
      S[sum] = i;
    } else {
      S[sum] = min(S[sum], i);
    }
  }
  cout << 1 << endl;
  int64_t T;
  cin >> T;
  for (int64_t t = 1; t <= T; ++t) {
    int64_t C;
    cin >> C;
    cout << S[C] << endl;
  }
  return 0;
}
