Orac is studying number theory, and he is interested in the properties of divisors.

For two positive integers a and b, a is a divisor of b if and only if there exists an integer c, such that a⋅ c=b.

For n ≥ 2, we will denote as f(n) the smallest positive divisor of n, except 1.

For example, f(7)=7,f(10)=2,f(35)=5.

For the fixed integer n, Orac decided to add f(n) to n. 

For example, if he had an integer n=5, the new value of n will be equal to 10. And if he had an integer n=6, n will be changed to 8.

Orac loved it so much, so he decided to repeat this operation several times.

Now, for two positive integers n and k, Orac asked you to add f(n) to n exactly k times (note that n will change after each operation, so f(n) may change too) and tell him the final value of n.

For example, if Orac gives you n=5 and k=2, at first you should add f(5)=5 to n=5, so your new value of n will be equal to n=10, after that, you should add f(10)=2 to 10, so your new (and the final!) value of n will be equal to 12.

Orac may ask you these queries many times.

Input

The first line of the input is a single integer t\ (1≤ t≤ 100): the number of times that Orac will ask you.

Each of the next t lines contains two positive integers n,k\ (2≤ n≤ 10^6, 1≤ k≤ 10^9), corresponding to a query by Orac.

It is guaranteed that the total sum of n is at most 10^6. 

Output

Print t lines, the i-th of them should contain the final value of n in the i-th query by Orac.

Example

Input


3
5 1
8 2
3 4


Output


10
12
12

Note

In the first query, n=5 and k=1. The divisors of 5 are 1 and 5, the smallest one except 1 is 5. Therefore, the only operation adds f(5)=5 to 5, and the result is 10.

In the second query, n=8 and k=2. The divisors of 8 are 1,2,4,8, where the smallest one except 1 is 2, then after one operation 8 turns into 8+(f(8)=2)=10. The divisors of 10 are 1,2,5,10, where the smallest one except 1 is 2, therefore the answer is 10+(f(10)=2)=12.

In the third query, n is changed as follows: 3 → 6 → 8 → 10 → 12.
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using ld = long double;
ll INF = LLONG_MAX;
using vi = vector<int>;
using vll = vector<ll>;
using pii = pair<int, int>;
namespace output {
void pr(int x) { cout << x; }
void pr(long x) { cout << x; }
void pr(ll x) { cout << x; }
void pr(unsigned x) { cout << x; }
void pr(unsigned long x) { cout << x; }
void pr(unsigned long long x) { cout << x; }
void pr(float x) { cout << x; }
void pr(double x) { cout << x; }
void pr(ld x) { cout << x; }
void pr(char x) { cout << x; }
void pr(const char* x) { cout << x; }
void pr(const string& x) { cout << x; }
void pr(bool x) { pr(x ? "true" : "false"); }
template <class T>
void pr(const complex<T>& x) {
  cout << x;
}
template <class T1, class T2>
void pr(const pair<T1, T2>& x);
template <class T>
void pr(const T& x);
template <class T, class... Ts>
void pr(const T& t, const Ts&... ts) {
  pr(t);
  pr(ts...);
}
template <class T1, class T2>
void pr(const pair<T1, T2>& x) {
  pr("{", x.f, ", ", x.s, "}");
}
template <class T>
void pr(const T& x) {
  pr("{");
  bool fst = 1;
  for (const auto& a : x) pr(!fst ? ", " : "", a), fst = 0;
  pr("}");
}
void print() { pr("\n"); }
template <class T, class... Ts>
void print(const T& t, const Ts&... ts) {
  pr(t);
  if (sizeof...(ts)) pr(" ");
  print(ts...);
}
}  // namespace output
using namespace output;
void solve() {
  ll N, K;
  cin >> N >> K;
  while (N % 2 != 0) {
    ll d = 2;
    while (N % d != 0) ++d;
    N += d;
    --K;
  }
  print(N + 2 * K);
}
int main() {
  ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  int T;
  cin >> T;
  for (int i = (0); i < (T); ++i) solve();
}
