#include <bits/stdc++.h>
using namespace std;
#define int long long
using i64 = long long;
const int N = 3e5 + 10;
int S[N], A[N], B[N];
int nx[N], val[N];
#define all(x) x.begin(), x.end()
void solve()
{
    int n;
    cin >> n;
    i64 ans = 0;
    vector<vector<int>> p(n);
    set<int> st;
    i64 SB = 0;
    vector<int> fa(n);
    iota(all(fa), 0);
    function<int(int x)> find = [&](int x)
    {
        return fa[x] == x ? x : fa[x] = find(fa[x]);
    };
    for (int i = 0, j = 0; i < n; i++)
    {
        int s, a, b;
        cin >> s >> a >> b;
        s = (s + ans) % n;
        S[i] = s, A[i] = a, B[i] = b;
        if (i)
        {
            while (j && S[i] != S[j])
                j = nx[j - 1];
            if (S[i] == S[j])
                j++;
            nx[i] = j;
        }
      
        // 上一个状态等价于nx[i-1]-1
        if (i && nx[i - 1] && S[i] == S[nx[i - 1]])
            fa[i - 1] = find(nx[i - 1] - 1);
        if (S[i] == S[0])
        {
            SB += B[i];
        }
        int p = i - 1;
        while (~p)
        {
            // 找到等价点的父节点
            if (S[p + 1] == S[i])
                p = find(p);
            // 不满足,p+1是长度，i-p-1
            if (S[p + 1] != S[i])
            {
                SB -= B[i - p - 1];
            }
            p = nx[p]-1;
        }
        ans += SB * A[i];
        cout << ans << endl;
    }
}
signed main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    int T = 1;
    while (T--)
    {
        solve();
    }
}