Problem: You are given two positive integers a and b. There are two possible operations: 

  1. multiply one of the numbers by some prime p; 
  2. divide one of the numbers on its prime factor p. 



What is the minimum number of operations required to obtain two integers having the same number of divisors? You are given several such pairs, you need to find the answer for each of them.

Input

The first line contains a single integer t (1 ≤ t ≤ 10^5) — the number of pairs of integers for which you are to find the answer.

Each of the next t lines contain two integers a_i and b_i (1 ≤ a_i, b_i ≤ 10^6).

Output

Output t lines — the i-th of them should contain the answer for the pair a_i, b_i.

Example

Input

8
9 10
100 17
220 70
17 19
4 18
32 20
100 32
224 385


Output

1
3
1
0
1
0
1
1

Note

These are the numbers with equal number of divisors, which are optimal to obtain in the sample test case: 

  * (27, 10), 4 divisors 
  * (100, 1156), 9 divisors 
  * (220, 140), 12 divisors 
  * (17, 19), 2 divisors 
  * (12, 18), 6 divisors 
  * (50, 32), 6 divisors 
  * (224, 1925), 12 divisors 



Note that there can be several optimal pairs of numbers.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
int sieve();
vector<int> numPF(long long x);
int fn(int current, long long with_me, int beforeCnt);
int get_block_differences(vector<int> x, vector<int> y);
void pre_process();
int get_number_index_in_block_array(int x);
int solve(int x, int y);
int make_array_of_different_divisors();
pair<vector<int>, int> different_blocks[1000];
map<int, int> different_divisors;
int no_of_different_blocks = 0;
int no_of_different_divisors = 0;
int no_of_different_primes = 0;
bool v[1000009];
long long primes[1000009];
int tmp[1000];
const int inf = 99999999;
int blocks_divisors_array[1000][1000];
int main() {
  pre_process();
  int tc;
  scanf("%d", &tc);
  for (int i = 0; i < tc; i++) {
    int x, y;
    scanf("%d %d", &x, &y);
    printf("%d\n", solve(x, y));
  }
  return 0;
}
int solve(int x, int y) {
  int x_index = get_number_index_in_block_array(x);
  int y_index = get_number_index_in_block_array(y);
  int ret = inf;
  for (map<int, int>::iterator it = different_divisors.begin();
       it != different_divisors.end(); it++) {
    pair<int, int> x = *it;
    int div_index = x.second;
    ret = min(ret, blocks_divisors_array[x_index][div_index] +
                       blocks_divisors_array[y_index][div_index]);
  }
  return ret;
}
void pre_process() {
  no_of_different_primes = sieve();
  no_of_different_blocks = fn(0, 1, 70);
  no_of_different_divisors = make_array_of_different_divisors();
  pair<vector<int>, int> x;
  x.first.push_back(0), x.second = 1;
  different_blocks[no_of_different_blocks++] = x;
  for (int i = 0; i < no_of_different_blocks; i++)
    for (int j = 0; j < no_of_different_divisors; j++)
      blocks_divisors_array[i][j] = inf;
  for (int i = 0; i < no_of_different_blocks; i++) {
    for (int j = 0; j < no_of_different_blocks; j++) {
      int x = get_block_differences(different_blocks[i].first,
                                    different_blocks[j].first);
      int div = different_blocks[j].second;
      int indx = different_divisors[div];
      blocks_divisors_array[i][indx] = min(blocks_divisors_array[i][indx], x);
    }
  }
}
int get_block_differences(vector<int> x, vector<int> y) {
  int i = x.size() - 1, j = y.size() - 1;
  int ret = 0;
  while (i >= 0 || j >= 0) {
    if (i < 0)
      ret += y[j];
    else if (j < 0)
      ret += x[i];
    else
      ret += abs(x[i] - y[j]);
    i--, j--;
  }
  return ret;
}
int get_number_index_in_block_array(int x) {
  vector<int> prime_dist = numPF(x);
  for (int i = 0; i < no_of_different_blocks; i++) {
    if (different_blocks[i].first == prime_dist) {
      return i;
    }
  }
  return -1;
}
int fn(int current, long long with_me, int beforeCnt) {
  int x = 60;
  if (with_me > 1) {
    pair<vector<int>, int> x;
    int div = 1;
    for (int i = 0; i < current; i++)
      x.first.push_back(tmp[i]), div = div * (tmp[i] + 1);
    sort(x.first.begin(), x.first.end());
    x.second = div;
    different_blocks[no_of_different_blocks++] = x;
  }
  if (current == no_of_different_primes ||
      with_me * primes[current] > 1000009 * x) {
    return 1;
  }
  long long local = 1;
  int ret = 0;
  for (int i = 1; i <= beforeCnt && with_me * local < 1000009 * x; i++) {
    local *= primes[current];
    if (with_me * local >= 1000009 * x) break;
    tmp[current] = i;
    ret += fn(current + 1, with_me * local, i);
  }
  return ret + 1;
}
int make_array_of_different_divisors() {
  set<int> divisors_set;
  for (int i = 0; i < no_of_different_blocks; i++)
    divisors_set.insert(different_blocks[i].second);
  divisors_set.insert(1);
  int j = 0;
  for (set<int>::iterator it = divisors_set.begin(); it != divisors_set.end();
       it++) {
    int x = *it;
    different_divisors[x] = j;
    j++;
  }
  return divisors_set.size();
}
int sieve() {
  memset(v, 0, sizeof(v));
  v[0] = v[1] = 1;
  long long i, j;
  for (i = 2; i * i < 1000009; i++) {
    if (!v[i]) {
      for (j = i * i; j < 1000009; j += i) {
        v[j] = 1;
      }
    }
  }
  int index = 0;
  for (i = 2; i < 1000009; i++) {
    if (!v[i]) {
      primes[index++] = i;
    }
  }
  return index;
}
vector<int> numPF(long long x) {
  int i = 0;
  vector<int> ret;
  if (x == 1) ret.push_back(0);
  long long pf = primes[i++];
  while (x != 1 && i < no_of_different_primes && x >= pf * pf) {
    int cnt = 0;
    while (x % pf == 0) {
      x /= pf;
      cnt++;
    }
    if (cnt) ret.push_back(cnt);
    pf = primes[i++];
  }
  if (x != 1) ret.push_back(1);
  sort(ret.begin(), ret.end());
  return ret;
}