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
N,K=map(int,input().split())
A=[int(input()) for i in range(N)]

for i in range(N):
    A[i]-=K

from itertools import accumulate
S=[0]+list(accumulate(A))

SORTED_S=sorted(set(S)|{-1<<63})
compression_dict={a: ind for ind, a in enumerate(SORTED_S)}

LEN=len(compression_dict)

BIT=[0]*(LEN+1) # 1-indexedなtree

def update(v,w): # index vにwを加える
    while v<=LEN:
        BIT[v]+=w
        v+=(v&(-v)) # 自分を含む大きなノードへ. たとえばv=3→v=4

def getvalue(v): # [1,v]の区間の和を求める
    ANS=0
    while v!=0:
        ANS+=BIT[v]
        v-=(v&(-v)) # 自分より小さい2ベキのノードへ. たとえばv=3→v=2へ
    return ANS

ANS=0
for i in range(N+1):
    ANS+=getvalue(compression_dict[S[i]])
    update(compression_dict[S[i]],1)

print(ANS)