Catherine received an array of integers as a gift for March 8. Eventually she grew bored with it, and she started calculated various useless characteristics for it. She succeeded to do it for each one she came up with. But when she came up with another one — xor of all pairwise sums of elements in the array, she realized that she couldn't compute it for a very large array, thus she asked for your help. Can you do it? Formally, you need to compute

$$$ (a_1 + a_2) ⊕ (a_1 + a_3) ⊕ … ⊕ (a_1 + a_n) \\\ ⊕ (a_2 + a_3) ⊕ … ⊕ (a_2 + a_n) \\\ … \\\ ⊕ (a_{n-1} + a_n) \\\ $$$

Here x ⊕ y is a bitwise XOR operation (i.e. x ^ y in many modern programming languages). You can read about it in Wikipedia: <https://en.wikipedia.org/wiki/Exclusive_or#Bitwise_operation>.

Input

The first line contains a single integer n (2 ≤ n ≤ 400 000) — the number of integers in the array.

The second line contains integers a_1, a_2, …, a_n (1 ≤ a_i ≤ 10^7).

Output

Print a single integer — xor of all pairwise sums of integers in the given array.

Examples

Input


2
1 2


Output


3

Input


3
1 2 3


Output


2

Note

In the first sample case there is only one sum 1 + 2 = 3.

In the second sample case there are three sums: 1 + 2 = 3, 1 + 3 = 4, 2 + 3 = 5. In binary they are represented as 011_2 ⊕ 100_2 ⊕ 101_2 = 010_2, thus the answer is 2.

⊕ is the bitwise xor operation. To define x ⊕ y, consider binary representations of integers x and y. We put the i-th bit of the result to be 1 when exactly one of the i-th bits of x and y is 1. Otherwise, the i-th bit of the result is put to be 0. For example, 0101_2   ⊕   0011_2 = 0110_2.
import io
import os

from collections import deque, defaultdict, Counter

from bisect import bisect_left, bisect_right


DEBUG = False


def solveBrute(N, A):
    ans = 0
    for i in range(N):
        for j in range(i + 1, N):
            ans ^= A[i] + A[j]
    return ans


def solve(N, A):
    B = max(A).bit_length()
    ans = 0
    for k in range(B + 1):
        # Count number of pairs with kth bit on (0 indexed)
        # For example if k==2, want pairs where lower 3 bits are between 100 and 111 inclusive
        # If we mask A to the lower 3 bits, we can find all pairs that sum to either 100 to 111 or overflowed to 1100 to 1111

        MOD = 1 << (k + 1)
        MASK = MOD - 1
        # Sort by x & MASK incrementally
        i = 0
        right = []
        for x in A:
            if (x >> k) & 1:
                right.append(x)
            else:
                A[i] = x
                i += 1
        assert N - i == len(right)
        A[i:] = right
        arr = [x & MASK for x in A]
        if DEBUG:
            assert arr == sorted(arr)

        numPairs = 0
        tlo = 1 << k
        thi = (1 << (k + 1)) - 1

        for targetLo, targetHi in [(tlo, thi), (MOD + tlo, MOD + thi)]:
            # Want to binary search for y such that targetLo <= x + y <= targetHi
            # But this TLE so walk the lo/hi pointers instead
            lo = N
            hi = N
            for i, x in enumerate(arr):
                lo = max(lo, i + 1)
                hi = max(hi, lo)
                while lo - 1 >= i + 1 and arr[lo - 1] >= targetLo - x:
                    lo -= 1
                while hi - 1 >= lo and arr[hi - 1] >= targetHi - x + 1:
                    hi -= 1
                numPairs += hi - lo

                if DEBUG:
                    assert lo == bisect_left(arr, targetLo - x, i + 1)
                    assert hi == bisect_right(arr, targetHi - x, lo)
                    # Check
                    for j, y in enumerate(arr):
                        cond = i < j and targetLo <= x + y <= targetHi
                        if lo <= j < hi:
                            assert cond
                        else:
                            assert not cond

        ans += (numPairs % 2) << k

    return ans


if DEBUG:
    import random

    random.seed(0)
    for i in range(100):
        A = [random.randint(1, 1000) for i in range(100)]
        N = len(A)
        ans1 = solveBrute(N, A)
        ans2 = solve(N, A)
        print(A, bin(ans1), bin(ans2))
        assert ans1 == ans2
else:
    if False:
        # Timing
        import random

        random.seed(0)
        A = [random.randint(1, 10 ** 7) for i in range(400000)]
        N = len(A)
        print(solve(N, A))


if __name__ == "__main__":
    (N,) = list(map(int, input().split()))
    A = list(map(int, input().split()))
    ans = solve(N, A)
    print(ans)
