Due to the coronavirus pandemic, city authorities obligated citizens to keep a social distance. The mayor of the city Semyon wants to light up Gluharniki park so that people could see each other even at night to keep the social distance.

The park is a rectangular table with n rows and m columns, where the cells of the table are squares, and the boundaries between the cells are streets. External borders are also streets. Every street has length 1. For example, park with n=m=2 has 12 streets.

You were assigned to develop a plan for lighting the park. You can put lanterns in the middle of the streets. The lamp lights two squares near it (or only one square if it stands on the border of the park).

<image> The park sizes are: n=4, m=5. The lighted squares are marked yellow. Please note that all streets have length 1. Lanterns are placed in the middle of the streets. In the picture not all the squares are lit.

Semyon wants to spend the least possible amount of money on lighting but also wants people throughout the park to keep a social distance. So he asks you to find the minimum number of lanterns that are required to light all the squares.

Input

The first line contains a single integer t (1 ≤ t ≤ 10^4) — the number of test cases in the input. Then t test cases follow.

Each test case is a line containing two integers n, m (1 ≤ n, m ≤ 10^4) — park sizes.

Output

Print t answers to the test cases. Each answer must be a single integer — the minimum number of lanterns that are required to light all the squares.

Example

Input


5
1 1
1 3
2 2
3 3
5 3


Output


1
2
2
5
8

Note

Possible optimal arrangement of the lanterns for the 2-nd test case of input data example: <image>

Possible optimal arrangement of the lanterns for the 3-rd test case of input data example: <image>
#include <bits/stdc++.h>
using namespace std;
inline long long int modadd(long long int n, long long int m,
                            long long int p = 1000000007) {
  return ((n + m) % p + p) % p;
}
inline long long int modsub(long long int n, long long int m,
                            long long int p = 1000000007) {
  return ((n - m + p) % p + p) % p;
}
inline long long int modpro(long long int n, long long int m,
                            long long int p = 1000000007) {
  return (((n % p) * (m % p)) % p + p) % p;
}
unsigned long long int powe(long long int first, long long int second) {
  unsigned long long int res = 1;
  while (second > 0) {
    if (second & 1) res = res * first;
    second = second >> 1;
    first = first * first;
  }
  return res;
}
long long int modpow(long long int first, long long int second,
                     long long int p = 1000000007) {
  long long int res = 1;
  while (second > 0) {
    if (second & 1) res = modpro(res, first, p);
    second = second >> 1;
    first = modpro(first, first, p);
  }
  return res;
}
inline long long int modInverse(long long int n, long long int p = 1000000007) {
  if (n == 1) return 1;
  return modpow(n, p - 2, p);
}
inline long long int moddiv(long long int n, long long int m,
                            long long int p = 1000000007) {
  return modpro(n, modInverse(m, p), p);
}
inline long long int modadd3(long long int first, long long int second,
                             long long int z, long long int p = 1000000007) {
  return modadd(modadd(first, second, p), z, p);
}
inline long long int modadd4(long long int first, long long int second,
                             long long int z, long long int w,
                             long long int p = 1000000007) {
  return modadd(modadd(first, second, p), modadd(z, w, p), p);
}
inline long long int modnCr(long long int fac[], int n, int r,
                            long long int p = 1000000007) {
  if (r == 0) return 1;
  return modpro(fac[n], modInverse(modpro(fac[r], fac[n - r], p), p), p);
}
template <typename T>
inline T max3(T first, T second, T z) {
  return max(max(first, second), z);
}
template <typename T>
inline T max4(T first, T second, T z, T w) {
  return max(max3(first, second, w), z);
}
template <typename T>
inline T min3(T first, T second, T z) {
  return min(min(first, second), z);
}
template <typename T>
inline T min4(T first, T second, T z, T w) {
  return min(min3(first, second, w), z);
}
template <typename T>
void printArr(T *arr, int s, int n) {
  for (int i = s; i <= n; i++) {
    cout << arr[i] << " ";
  }
  cout << endl;
}
int a[1005];
int main() {
  std::ios::sync_with_stdio(false);
  cin.tie(NULL);
  int erer = 1;
  cin >> erer;
  for (int erer2 = (1); erer2 < (erer + 1); erer2++) {
    int n, m;
    cin >> n >> m;
    if (n % 2 == 0) {
      cout << n / 2 * m << endl;
    } else {
      cout << n / 2 * m + (m + 1) / 2 << endl;
    }
  }
  return 0;
}
