You are given an integer sequence a_1, a_2, ..., a_n.

Find the number of pairs of indices (l, r) (1 ≤ l ≤ r ≤ n) such that the value of median of a_l, a_{l+1}, ..., a_r is exactly the given number m.

The median of a sequence is the value of an element which is in the middle of the sequence after sorting it in non-decreasing order. If the length of the sequence is even, the left of two middle elements is used.

For example, if a=[4, 2, 7, 5] then its median is 4 since after sorting the sequence, it will look like [2, 4, 5, 7] and the left of two middle elements is equal to 4. The median of [7, 1, 2, 9, 6] equals 6 since after sorting, the value 6 will be in the middle of the sequence.

Write a program to find the number of pairs of indices (l, r) (1 ≤ l ≤ r ≤ n) such that the value of median of a_l, a_{l+1}, ..., a_r is exactly the given number m.

Input

The first line contains integers n and m (1 ≤ n,m ≤ 2⋅10^5) — the length of the given sequence and the required value of the median.

The second line contains an integer sequence a_1, a_2, ..., a_n (1 ≤ a_i ≤ 2⋅10^5).

Output

Print the required number.

Examples

Input

5 4
1 4 5 60 4


Output

8


Input

3 1
1 1 1


Output

6


Input

15 2
1 2 3 1 2 3 1 2 3 1 2 3 1 2 3


Output

97

Note

In the first example, the suitable pairs of indices are: (1, 3), (1, 4), (1, 5), (2, 2), (2, 3), (2, 5), (4, 5) and (5, 5).
import java.io.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.StringTokenizer;
import java.io.IOException;
import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.io.InputStream;

/**
 * Built using CHelper plug-in
 * Actual solution is at the top
 *
 * @author kessido
 */
public class Main {
    public static void main(String[] args) {
        InputStream inputStream = System.in;
        OutputStream outputStream = System.out;
        InputReader in = new InputReader(inputStream);
        PrintWriter out = new PrintWriter(outputStream);
        E2MedianOnSegmentsGeneralCaseEdition solver = new E2MedianOnSegmentsGeneralCaseEdition();
        solver.solve(1, in, out);
        out.close();
    }

    static class E2MedianOnSegmentsGeneralCaseEdition {
        public void solve(int testNumber, InputReader in, PrintWriter out) {
            int n = in.NextInt();
            int m = in.NextInt();
            int[] a = in.NextIntArray(n);
            long ansM = getNumberOfSegmentWithMajorityOfElementLessOrEqaulToM(a, m);
            long ansM_1 = getNumberOfSegmentWithMajorityOfElementLessOrEqaulToM(a, m - 1);
            out.println(ansM - ansM_1);
        }

        private long getNumberOfSegmentWithMajorityOfElementLessOrEqaulToM(int[] a, int m) {
            int shift = a.length;
            FenwickTree fw = new FenwickTree(2 * a.length + 1);
            int cur = 0;
            long res = 0;
            for (int i : a) {
                fw.addToIndex(shift + cur, 1);
                if (i > m) {
                    cur--;
                } else cur++;
                // need cur - x >= 0
                // x > cur
                res += fw.getSum(shift + cur);
            }
            return res;
        }

    }

    static interface IPrefixSum extends ArraysInterface.ILongArray {
    }

    static interface IFindFirstBiggerIndex<T> {
    }

    static class InputReader {
        BufferedReader reader;
        StringTokenizer tokenizer;

        public InputReader(InputStream stream) {
            reader = new BufferedReader(new InputStreamReader(stream), 32768);
            tokenizer = null;
        }

        public String next() {
            while (tokenizer == null || !tokenizer.hasMoreTokens()) {
                try {
                    tokenizer = new StringTokenizer(reader.readLine(), " \t\n\r\f,");
                } catch (IOException e) {
                    throw new RuntimeException(e);
                }
            }
            return tokenizer.nextToken();
        }

        public int NextInt() {
            return Integer.parseInt(next());
        }

        public int[] NextIntArray(int n) {
            return NextIntArray(n, 0);
        }

        public int[] NextIntArray(int n, int offset) {
            int[] a = new int[n];
            for (int i = 0; i < n; i++) {
                a[i] = NextInt() + offset;
            }
            return a;
        }

    }

    static class FenwickTree implements IPrefixSum, IFindFirstBiggerIndex<Long> {
        public final int length;
        private final long[] a;
        private long sum = 0;

        public FenwickTree(final int length) {
            assert length != 0;
            this.length = length;
            this.a = new long[length + 1];
        }

        public FenwickTree(final long[] array) {
            this(array.length);
            for (int i = 0; i < length; i++) {
                a[i + 1] = a[i] + array[i];
            }
            sum = a[length];
            for (int i = length; i > 0; i--) {
                int ind = i - ((i) & (-i)); //LSB
                a[i] -= a[ind];
            }
        }


        public void addToIndex(int index, final long value) {
            index++;
            while (index <= length) {
                a[index] += value;
                index += (index & (-index)); //LSB
            }
            sum += value;
        }


        public long getSum(int index) {
            long sum = 0;
            index++;
            while (index > 0) {
                sum += a[index];
                index -= ((index) & (-index)); //LSB
            }
            return sum;
        }

    }

    static class ArraysInterface {
        public interface ILongArray {
        }

    }
}

