# ビット演算なのでビット桁ごとに計算、ORなので1があるかどうか
# 入力例1の2の0乗桁は、1,1,0、つまり1のない区間は1，ある区間はN*(N+1)//2-1=5

N = int(input())
A = list(map(int, input().split()))

ans = 0
for d in range(0, 30):
    zero_segment = 0
    zero_count = 0
    for a in A:
        if (a>>d)&1 == 0:
            zero_count += 1
        else:
            zero_segment += zero_count*(zero_count+1)//2
            zero_count = 0
    zero_segment += zero_count*(zero_count+1)//2
    calc = (1<<d)*(N*(N+1)//2-zero_segment)
    ans += calc
    #print('d', d, 'zero_segment', zero_segment, 'calc', calc)
print(ans)