Sum of XOR Functions, Contribution 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); }
// -----------------------------------------------------------------------------
void solve()
{
    int n;
    cin >> n;
    vi a(n);
    for (int i = 0; i < n; i++)
    {
        cin >> a[i];
    }
    int sum = 0;
    // sum of (r-l+1) = r * (cnt[different bit]) - summation of idxs(where different bit is there)
    for (int i = 0; i < 31; i++)
    {
        vi cnt(2, 0), sumidx(2, 0);
        cnt[0] = 1; // else starting from 1 wont be counted
        int cur = 0, pref = 0;
        for (int j = 0; j < n; j++)
        {
            int bit = (a[j] >> i) & 1;
            pref ^= bit; // prefix bit
            int contriR = (cnt[pref ^ 1] * (j + 1)) % mod2;
            cur = (cur + contriR - sumidx[pref ^ 1] + mod2) % mod2;
            cnt[pref]++;
            sumidx[pref] = (sumidx[pref] + (j + 1)) % mod2;
        }
        sum = (sum + ((1LL << i) * cur) % mod2) % mod2;
    }
    cout << sum << endl;
    return;
}

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