codeforces#P1879D. Sum of XOR Functions

    ID: 34112 远端评测题 2000ms 256MiB 尝试: 0 已通过: 0 难度: 5 上传者: 标签>bitmaskscombinatoricsdivide and conquerdpmath*1700

Sum of XOR Functions

Description

You are given an array $a$ of length $n$ consisting of non-negative integers.

You have to calculate the value of $\sum_{l=1}^{n} \sum_{r=l}^{n} f(l, r) \cdot (r - l + 1)$, where $f(l, r)$ is $a_l \oplus a_{l+1} \oplus \dots \oplus a_{r-1} \oplus a_r$ (the character $\oplus$ denotes bitwise XOR).

Since the answer can be very large, print it modulo $998244353$.

The first line contains one integer $n$ ($1 \le n \le 3 \cdot 10^5$) — the length of the array $a$.

The second line contains $n$ integers $a_1, a_2, \dots, a_n$ ($0 \le a_i \le 10^9)$.

Print the one integer — the value of $\sum_{l=1}^{n} \sum_{r=l}^{n} f(l, r) \cdot (r - l + 1)$, taken modulo $998244353$.

Input

The first line contains one integer $n$ ($1 \le n \le 3 \cdot 10^5$) — the length of the array $a$.

The second line contains $n$ integers $a_1, a_2, \dots, a_n$ ($0 \le a_i \le 10^9)$.

Output

Print the one integer — the value of $\sum_{l=1}^{n} \sum_{r=l}^{n} f(l, r) \cdot (r - l + 1)$, taken modulo $998244353$.

3
1 3 2
4
39 68 31 80
7
313539461 779847196 221612534 488613315 633203958 394620685 761188160
12
1337
257421502

Note

In the first example, the answer is equal to $f(1, 1) + 2 \cdot f(1, 2) + 3 \cdot f(1, 3) + f(2, 2) + 2 \cdot f(2, 3) + f(3, 3) = $ $= 1 + 2 \cdot 2 + 3 \cdot 0 + 3 + 2 \cdot 1 + 2 = 12$.