You are given an integer sequence of length N, a = {a_1, a_2, …, a_N}, and an integer K.

a has N(N+1)/2 non-empty contiguous subsequences, {a_l, a_{l+1}, …, a_r} (1 ≤ l ≤ r ≤ N). Among them, how many have an arithmetic mean that is greater than or equal to K?

Constraints

* All input values are integers.
* 1 ≤ N ≤ 2 \times 10^5
* 1 ≤ K ≤ 10^9
* 1 ≤ a_i ≤ 10^9

Input

Input is given from Standard Input in the following format:


N K
a_1
a_2
:
a_N


Output

Print the number of the non-empty contiguous subsequences with an arithmetic mean that is greater than or equal to K.

Examples

Input

3 6
7
5
7


Output

5


Input

1 2
1


Output

0


Input

7 26
10
20
30
40
30
20
10


Output

13
class BIT():

    def __init__(self, n):
        '''
        n = 要素数
        添字は i = 0 ~ n-1 となる
        '''
        self.n = n
        self.bit = [0] * (n + 1)

    def add(self, i, x):
        '''i番目の要素にxを加算する'''
        i = i + 1
        while i <= self.n:
            self.bit[i] += x
            i += i & -i

    def _sum(self, i):
        s = 0
        while i > 0:
            s += self.bit[i]
            i -= i & -i
        return s

    def get(self, i, j):
        '''[i, j)の和を求める'''
        return self._sum(j) - self._sum(i)


n, k = map(int, input().split())
a = [int(input()) for _ in range(n)]

#いい感じに式変形した結果の配列をつくる
#cumsum_a[i] <= cumsum_a[j]を満たすi<jの組の個数が答えになる
cumsum_a = [0]*(n+1)
for i in range(n):
    cumsum_a[i+1] = cumsum_a[i] + a[i]
for i in range(n+1):
    cumsum_a[i] = cumsum_a[i] - k*i

#座標圧縮したリストを返す
def compress(list1):
    list2 = sorted(set(list1))
    memo = {value : index for index, value in enumerate(list2)}
    for i in range(len(list1)):
        list1[i] = memo[list1[i]]
    return list1

cumsum_a = compress(cumsum_a)

bit = BIT(n+1)
ans = 0
for i in range(n+1):
    ans += bit.get(0, cumsum_a[i] + 1)
    bit.add(cumsum_a[i], 1)
print(ans)