XOR SUM 4, Contributuion Technique

#include <bits/stdc++.h>
using namespace std;
#define int long long
#define all(x) (x).begin(), (x).end()
typedef vector<int> vi;
typedef vector<vi> vvi;
typedef vector<pair<int, int>> vpi;
typedef pair<int, int> pi;
#define f first
#define s second
#define pb push_back
#define endl "\n"
#define yes cout << "YES" << endl
#define no cout << "NO" << endl
#define init(x, a) memset(x, a, sizeof(x))
const int mod1 = 1e9 + 7, mod2 = 998244353, INF = 2e18, N = 2e5 + 5, L = 19;
int gcd(int a, int b) { return b == 0 ? a : gcd(b, a % b); }
// -----------------------------------------------------------------------------
int modadd(int a, int b, int mod = (1e9 + 7))
{
    return ((a % mod) + (b % mod)) % mod;
}
int modmul(int a, int b, int mod = (1e9 + 7))
{
    return ((a % mod) * (b % mod)) % mod;
}
int modsub(int a, int b, int mod = (1e9 + 7))
{
    return ((a % mod) - (b % mod) + mod) % mod;
}
int binpow(int a, int b, int mod = (1e9 + 7))
{
    int res = 1;
    while (b > 0)
    {
        if (b & 1)
        {
            res = (res * a) % mod;
        }
        a = (a * a) % mod;
        b >>= 1;
    }
    return res;
}
int modinv(int a, int mod = (1e9 + 7))
{
    return binpow(a, mod - 2, mod);
}
void solve()
{
    int n;
    cin >> n;
    vi a(n);
    for (int i = 0; i < n; i++)
    {
        cin >> a[i];
    }
    int sum = 0;
    for (int b = 0; b < 61; b++)
    {
        vi cnt(2, 0);
        // cnt[0] = 1;
        int pref = 0, curr = 0;
        for (int i = 0; i < n; i++)
        {
            int bit = (a[i] >> b) & 1;
            pref ^= bit;
            curr = modadd(curr, cnt[bit ^ 1]);
            cnt[bit] = modadd(cnt[bit], 1);
        }
        int temp = modmul(curr, 1LL << b);
        sum = modadd(temp, sum);
    }
    cout << sum << endl;
    return;
}

signed main()
{
    // __START__;
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);
    int t = 1;
    // cin >> t;
    while (t--)
    {
        solve();
    }
    // __END__;
    return 0;
}