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.BufferedReader;
import java.io.InputStreamReader;
import java.io.InputStream;

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

    static class E2MedianOnSegmentsGeneralCaseEdition {
        int n;
        int m;
        int[] arr;

        public void readInput(Scanner sc) {
            n = sc.nextInt();
            m = sc.nextInt();
            arr = new int[n];
            for (int i = 0; i < n; i++)
                arr[i] = sc.nextInt();
        }

        public void solve(int testNumber, Scanner sc, PrintWriter pw) {
            int tc = 1;
            while (tc-- > 0) {
                readInput(sc);
                pw.print(subArraysOfMedian(m) - subArraysOfMedian(m + 1));
            }
        }

        private long subArraysOfMedian(int m) {
            FenwickTree ft = new FenwickTree(3 * n);
            ft.point_update(n + 1, 1);
            int prefix = 0;
            long ans = 0;
            for (int i = n - 1; i >= 0; i--) {
                prefix += arr[i] >= m ? 1 : -1;
                ans += ft.rsq(n + 1 + prefix - 1);
                ft.point_update(prefix + n + 1, 1);
            }
            return ans;
        }

        public class FenwickTree {
            int n;
            int[] ft;

            FenwickTree(int size) {
                n = size;
                ft = new int[n + 1];
            }

            long rsq(int b) {
                long sum = 0;
                while (b > 0) {
                    sum += ft[b];
                    b -= b & -b;
                }        //min?
                return sum;
            }

            void point_update(int k, int val) {
                while (k <= n) {
                    ft[k] += val;
                    k += k & -k;
                }        //min?
            }

        }

    }

    static class Scanner {
        StringTokenizer st;
        BufferedReader br;

        public Scanner(InputStream s) {
            br = new BufferedReader(new InputStreamReader(s));
        }

        public String next() {
            try {
                while (st == null || !st.hasMoreTokens())
                    st = new StringTokenizer(br.readLine());
                return st.nextToken();
            } catch (Exception e) {
                throw new RuntimeException(e);
            }
        }

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

    }
}

