It is the easy version of the problem. The only difference is that in this version n = 1.

In the cinema seats can be represented as the table with n rows and m columns. The rows are numbered with integers from 1 to n. The seats in each row are numbered with consecutive integers from left to right: in the k-th row from m (k - 1) + 1 to m k for all rows 1 ≤ k ≤ n.

1| 2| ⋅⋅⋅| m - 1| m  
---|---|---|---|---  
m + 1| m + 2| ⋅⋅⋅| 2 m - 1| 2 m  
2m + 1| 2m + 2| ⋅⋅⋅| 3 m - 1| 3 m  
\vdots| \vdots| \ddots| \vdots| \vdots  
m (n - 1) + 1| m (n - 1) + 2| ⋅⋅⋅| n m - 1| n m  
The table with seats indices

There are nm people who want to go to the cinema to watch a new film. They are numbered with integers from 1 to nm. You should give exactly one seat to each person.

It is known, that in this cinema as lower seat index you have as better you can see everything happening on the screen. i-th person has the level of sight a_i. Let's define s_i as the seat index, that will be given to i-th person. You want to give better places for people with lower sight levels, so for any two people i, j such that a_i < a_j it should be satisfied that s_i < s_j.

After you will give seats to all people they will start coming to their seats. In the order from 1 to nm, each person will enter the hall and sit in their seat. To get to their place, the person will go to their seat's row and start moving from the first seat in this row to theirs from left to right. While moving some places will be free, some will be occupied with people already seated. The inconvenience of the person is equal to the number of occupied seats he or she will go through.

Let's consider an example: m = 5, the person has the seat 4 in the first row, the seats 1, 3, 5 in the first row are already occupied, the seats 2 and 4 are free. The inconvenience of this person will be 2, because he will go through occupied seats 1 and 3.

Find the minimal total inconvenience (the sum of inconveniences of all people), that is possible to have by giving places for all people (all conditions should be satisfied).

Input

The input consists of multiple test cases. The first line contains a single integer t (1 ≤ t ≤ 100) — the number of test cases. Description of the test cases follows.

The first line of each test case contains two integers n and m (n = 1, 1 ≤ m ≤ 300) — the number of rows and places in each row respectively.

The second line of each test case contains n ⋅ m integers a_1, a_2, …, a_{n ⋅ m} (1 ≤ a_i ≤ 10^9), where a_i is the sight level of i-th person.

It's guaranteed that the sum of n ⋅ m over all test cases does not exceed 10^5.

Output

For each test case print a single integer — the minimal total inconvenience that can be achieved.

Example

Input


4
1 3
1 2 3
1 5
2 1 5 3 3
1 2
2 1
1 6
2 3 2 1 1 1


Output


3
6
0
1

Note

In the first test case, there is a single way to arrange people, because all sight levels are distinct. The first person will sit on the first seat, the second person will sit on the second place, the third person will sit on the third place. So inconvenience of the first person will be 0, inconvenience of the second person will be 1 and inconvenience of the third person will be 2. The total inconvenience is 0 + 1 + 2 = 3.

In the second test case, people should sit as follows: s_1 = 2, s_2 = 1, s_3 = 5, s_4 = 4, s_5 = 3. The total inconvenience will be 6.
import java.io.*;
import java.util.*;

import static java.lang.Math.*;
import static java.util.Arrays.*;

public class cf1566d1 {

    public static void main(String[] args) throws IOException {
        int t = ri();
        while (t --> 0) {
            int n = rni(), m = ni(), a[] = ria(n * m), awi[][] = new int[n * m][2];
            for (int i = 0; i < n * m; ++i) {
                awi[i][0] = a[i];
                awi[i][1] = i;
            }
            sort(awi, (x, y) -> x[0] == y[0] ? y[1] - x[1] : x[0] - y[0]);
            int order[] = new int[n * m], cur = 0, ans = 0;
            for (int i = 0; i < n * m; ++i) {
                order[awi[i][1]] = cur++;
            }
            bit bit = new bit(n * m, Integer::sum);
            for (int i = 0; i < n * m; ++i) {
                ans += bit.qry(order[i]);
                bit.upd(order[i], 1);
            }
            prln(ans);
        }
        close();
    }

    @FunctionalInterface
    interface IntOperator {
        int merge(int a, int b);
    }

    static class bit {
        IntOperator op;
        int n, bit[];

        bit(int size, IntOperator operator) {
            bit = new int[n = size + 1];
            op = operator;
        }

        bit(int[] arr, IntOperator operator) {
            bit = new int[(n = arr.length + 1)];
            op = operator;
            for (int i = 0; i < n - 1; ++i) upd(i, arr[i]);
        }

        void upd(int i, int x) {
            ++i;
            while (i < n) {
                bit[i] = op.merge(bit[i], x);
                i += i & (-i);
            }
        }

        int qry(int i) {
            int ans = 0;
            while (i > 0) {
                ans = op.merge(ans, bit[i]);
                i -= i & (-i);
            }
            return ans;
        }
    }

    static BufferedReader __i = new BufferedReader(new InputStreamReader(System.in));
    static PrintWriter __o = new PrintWriter(new OutputStreamWriter(System.out));
    static StringTokenizer input;
    static Random __r = new Random();

    // references
    // IBIG = 1e9 + 7
    // IMAX ~= 2e9
    // LMAX ~= 9e18

    // constants
    static final int IBIG = 1000000007;
    static final int IMAX = 2147483647;
    static final long LMAX = 9223372036854775807L;
    // math util
    static int minof(int a, int b, int c) {return min(a, min(b, c));}
    static int minof(int... x) {if (x.length == 1) return x[0]; if (x.length == 2) return min(x[0], x[1]); if (x.length == 3) return min(x[0], min(x[1], x[2])); int min = x[0]; for (int i = 1; i < x.length; ++i) if (x[i] < min) min = x[i]; return min;}
    static long minof(long a, long b, long c) {return min(a, min(b, c));}
    static long minof(long... x) {if (x.length == 1) return x[0]; if (x.length == 2) return min(x[0], x[1]); if (x.length == 3) return min(x[0], min(x[1], x[2])); long min = x[0]; for (int i = 1; i < x.length; ++i) if (x[i] < min) min = x[i]; return min;}
    static int maxof(int a, int b, int c) {return max(a, max(b, c));}
    static int maxof(int... x) {if (x.length == 1) return x[0]; if (x.length == 2) return max(x[0], x[1]); if (x.length == 3) return max(x[0], max(x[1], x[2])); int max = x[0]; for (int i = 1; i < x.length; ++i) if (x[i] > max) max = x[i]; return max;}
    static long maxof(long a, long b, long c) {return max(a, max(b, c));}
    static long maxof(long... x) {if (x.length == 1) return x[0]; if (x.length == 2) return max(x[0], x[1]); if (x.length == 3) return max(x[0], max(x[1], x[2])); long max = x[0]; for (int i = 1; i < x.length; ++i) if (x[i] > max) max = x[i]; return max;}
    static int powi(int a, int b) {if (a == 0) return 0; int ans = 1; while (b > 0) {if ((b & 1) > 0) ans *= a; a *= a; b >>= 1;} return ans;}
    static long powl(long a, int b) {if (a == 0) return 0; long ans = 1; while (b > 0) {if ((b & 1) > 0) ans *= a; a *= a; b >>= 1;} return ans;}
    static int fli(double d) {return (int) d;}
    static int cei(double d) {return (int) ceil(d);}
    static long fll(double d) {return (long) d;}
    static long cel(double d) {return (long) ceil(d);}
    static int gcd(int a, int b) {return b == 0 ? a : gcd(b, a % b);}
    static long gcd(long a, long b) {return b == 0 ? a : gcd(b, a % b);}
    static int[] exgcd(int a, int b) {if (b == 0) return new int[] {1, 0}; int[] y = exgcd(b, a % b); return new int[] {y[1], y[0] - y[1] * (a / b)};}
    static long[] exgcd(long a, long b) {if (b == 0) return new long[] {1, 0}; long[] y = exgcd(b, a % b); return new long[] {y[1], y[0] - y[1] * (a / b)};}
    static int randInt(int min, int max) {return __r.nextInt(max - min + 1) + min;}
    static long mix(long x) {x += 0x9e3779b97f4a7c15L; x = (x ^ (x >> 30)) * 0xbf58476d1ce4e5b9L; x = (x ^ (x >> 27)) * 0x94d049bb133111ebL; return x ^ (x >> 31);}
    // array util
    static void reverse(int[] a) {for (int i = 0, n = a.length, half = n / 2; i < half; ++i) {int swap = a[i]; a[i] = a[n - i - 1]; a[n - i - 1] = swap;}}
    static void reverse(long[] a) {for (int i = 0, n = a.length, half = n / 2; i < half; ++i) {long swap = a[i]; a[i] = a[n - i - 1]; a[n - i - 1] = swap;}}
    static void reverse(double[] a) {for (int i = 0, n = a.length, half = n / 2; i < half; ++i) {double swap = a[i]; a[i] = a[n - i - 1]; a[n - i - 1] = swap;}}
    static void reverse(char[] a) {for (int i = 0, n = a.length, half = n / 2; i < half; ++i) {char swap = a[i]; a[i] = a[n - i - 1]; a[n - i - 1] = swap;}}
    static void shuffle(int[] a) {int n = a.length - 1; for (int i = 0; i < n; ++i) {int ind = randInt(i, n); int swap = a[i]; a[i] = a[ind]; a[ind] = swap;}}
    static void shuffle(long[] a) {int n = a.length - 1; for (int i = 0; i < n; ++i) {int ind = randInt(i, n); long swap = a[i]; a[i] = a[ind]; a[ind] = swap;}}
    static void shuffle(double[] a) {int n = a.length - 1; for (int i = 0; i < n; ++i) {int ind = randInt(i, n); double swap = a[i]; a[i] = a[ind]; a[ind] = swap;}}
    static void rsort(int[] a) {shuffle(a); sort(a);}
    static void rsort(long[] a) {shuffle(a); sort(a);}
    static void rsort(double[] a) {shuffle(a); sort(a);}
    static int[] copy(int[] a) {int[] ans = new int[a.length]; for (int i = 0; i < a.length; ++i) ans[i] = a[i]; return ans;}
    static long[] copy(long[] a) {long[] ans = new long[a.length]; for (int i = 0; i < a.length; ++i) ans[i] = a[i]; return ans;}
    static double[] copy(double[] a) {double[] ans = new double[a.length]; for (int i = 0; i < a.length; ++i) ans[i] = a[i]; return ans;}
    static char[] copy(char[] a) {char[] ans = new char[a.length]; for (int i = 0; i < a.length; ++i) ans[i] = a[i]; return ans;}
    // input
    static void r() throws IOException {input = new StringTokenizer(rline());}
    static int ri() throws IOException {return Integer.parseInt(rline());}
    static long rl() throws IOException {return Long.parseLong(rline());}
    static double rd() throws IOException {return Double.parseDouble(rline());}
    static int[] ria(int n) throws IOException {int[] a = new int[n]; r(); for (int i = 0; i < n; ++i) a[i] = ni(); return a;}
    static void ria(int[] a) throws IOException {int n = a.length; r(); for (int i = 0; i < n; ++i) a[i] = ni();}
    static int[] riam1(int n) throws IOException {int[] a = new int[n]; r(); for (int i = 0; i < n; ++i) a[i] = ni() - 1; return a;}
    static void riam1(int[] a) throws IOException {int n = a.length; r(); for (int i = 0; i < n; ++i) a[i] = ni() - 1;}
    static long[] rla(int n) throws IOException {long[] a = new long[n]; r(); for (int i = 0; i < n; ++i) a[i] = nl(); return a;}
    static void rla(long[] a) throws IOException {int n = a.length; r(); for (int i = 0; i < n; ++i) a[i] = nl();}
    static double[] rda(int n) throws IOException {double[] a = new double[n]; r(); for (int i = 0; i < n; ++i) a[i] = nd(); return a;}
    static void rda(double[] a) throws IOException {int n = a.length; r(); for (int i = 0; i < n; ++i) a[i] = nd();}
    static char[] rcha() throws IOException {return rline().toCharArray();}
    static void rcha(char[] a) throws IOException {int n = a.length, i = 0; for (char c : rline().toCharArray()) a[i++] = c;}
    static String rline() throws IOException {return __i.readLine();}
    static String n() {return input.nextToken();}
    static int rni() throws IOException {r(); return ni();}
    static int ni() {return Integer.parseInt(n());}
    static long rnl() throws IOException {r(); return nl();}
    static long nl() {return Long.parseLong(n());}
    static double rnd() throws IOException {r(); return nd();}
    static double nd() {return Double.parseDouble(n());}
    // output
    static void pr(int i) {__o.print(i);}
    static void prln(int i) {__o.println(i);}
    static void pr(long l) {__o.print(l);}
    static void prln(long l) {__o.println(l);}
    static void pr(double d) {__o.print(d);}
    static void prln(double d) {__o.println(d);}
    static void pr(char c) {__o.print(c);}
    static void prln(char c) {__o.println(c);}
    static void pr(char[] s) {__o.print(new String(s));}
    static void prln(char[] s) {__o.println(new String(s));}
    static void pr(String s) {__o.print(s);}
    static void prln(String s) {__o.println(s);}
    static void pr(Object o) {__o.print(o);}
    static void prln(Object o) {__o.println(o);}
    static void prln() {__o.println();}
    static void pryes() {prln("yes");}
    static void pry() {prln("Yes");}
    static void prY() {prln("YES");}
    static void prno() {prln("no");}
    static void prn() {prln("No");}
    static void prN() {prln("NO");}
    static boolean pryesno(boolean b) {prln(b ? "yes" : "no"); return b;};
    static boolean pryn(boolean b) {prln(b ? "Yes" : "No"); return b;}
    static boolean prYN(boolean b) {prln(b ? "YES" : "NO"); return b;}
    static void prln(int... a) {for (int i = 0, len = a.length - 1; i < len; pr(a[i]), pr(' '), ++i); if (a.length > 0) prln(a[a.length - 1]); else prln();}
    static void prln(long... a) {for (int i = 0, len = a.length - 1; i < len; pr(a[i]), pr(' '), ++i); if (a.length > 0) prln(a[a.length - 1]); else prln();}
    static void prln(double... a) {for (int i = 0, len = a.length - 1; i < len; pr(a[i]), pr(' '), ++i); if (a.length > 0) prln(a[a.length - 1]); else prln();}
    static <T> void prln(Collection<T> c) {int n = c.size() - 1; Iterator<T> iter = c.iterator(); for (int i = 0; i < n; pr(iter.next()), pr(' '), ++i); if (n >= 0) prln(iter.next()); else prln();}
    static void h() {prln("hlfd"); flush();}
    static void flush() {__o.flush();}
    static void close() {__o.close();}
}