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
import sys,bisect,math
input = sys.stdin.readline

# i が与えられたとき a1~aiまでの和を計算する( O(log n) )
def bitsum(bit,i):
    s = 0
    while i > 0:
        s += bit[i]
        i -= i&(-i)
    return s
    
# i と x が与えられたとき ai += x とする ( O(log n) )
def bitadd(bit,i,x):
    while i <= n:
        bit[i] += x
        i += i&(-i)
    return bit

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

#b[j+1]-b[i]:a[i]~a[j]
b = [0]
for i in range(n):
    b.append(b[-1]+a[i])

#c[i]:b[i]-k*i
c = [0]*(n+1)
for i in range(n+1):
    c[i] = b[i]-k*i

d = sorted(c)
for i in range(n+1):
    k = bisect.bisect_left(d,c[i])
    d[k] = d[k]-0.1
    c[i] = k+1

n = n+1
n_ = int(math.log2(n)) + 1
bit = [0] * (2**n_+1)
ass = 0
for j in range(n):
    ass += j-bitsum(bit,c[j])
    #bitsum(bit,a[j]): j-1番目までの中でa[j]以下のもの
    bit = bitadd(bit,c[j],1)


print(n*(n-1)//2-ass)