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).
L1=list(map(int, input().split()))
numList=list(map(int, input().split()))
length=L1[0]
m=L1[1]
def greaterCount(numList,m):
    countDic={0:1}
    sum=0
    total=0
    rem=0
    for number in numList:
        if number>=m:
            sum+=1
            rem+=countDic[sum-1]
            total+=rem
        else:
            sum-=1
            if sum in countDic:
                rem-=countDic[sum]
            total+=rem

        if sum in countDic:
            countDic[sum] += 1
        else:
            countDic[sum] = 1

    #print("m=", m, "number=", number, "sum=", sum, "total=", total, "rem=", rem, "countDic=", countDic)
    return total
print(greaterCount(numList,m)-greaterCount(numList,m+1))