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 copy

class BinaryIndexedTree():
    def __init__(self, max_n):
        self.size = max_n + 1
        self.tree = [0] * self.size
        self.depth = self.size.bit_length()
 
    def initialize(self, seq):
        for i, x in enumerate(seq[1:], 1):
            self.tree[i] += x
            j = i + (i & (-i))
            if j < self.size:
                self.tree[j] += self.tree[i]
 
    def __repr__(self):
        return self.tree.__repr__()
 
    def get_sum(self, i):
        s = 0
        while i:
            s += self.tree[i]
            i -= i & -i
        return s
 
    def add(self, i, x):
        while i < self.size:
            self.tree[i] += x
            i += i & -i
 
    def find_kth_element(self, k):
        x, sx = 0, 0
        dx = (1 << self.depth)
        for i in range(self.depth - 1, -1, -1):
            dx = (1 << i)
            if x + dx >= self.size:
                continue
            y = x + dx
            sy = sx + self.tree[y]
            if sy < k:
                x, sx = y, sy
        return x + 1

n,k = map(int,input().split())
a = [int(input()) for _ in range(n)]
ssum = [0]*(n+1)
for i in range(1,n+1):
    ssum[i] = ssum[i-1] + a[i-1]

r = [0]*(n+1)
for i in range(n+1):
    r[i] = ssum[i]-i*k

r2 = copy.copy(r)
r2.sort()

d = dict()
cnt = 1
d[r2[0]] = 1
for i in range(1,n+1):
    if r2[i] != r2[i-1]:
        cnt += 1
    d[r2[i]] = cnt
#print(d)

bit = BinaryIndexedTree(2*10**5+5)
ans = 0

for i in range(n,-1,-1):
    t = d[r[i]]
    ans += bit.get_sum(2*10**5+2)-bit.get_sum(t-1)
    bit.add(t, 1)

print(ans)