#include <bits/stdc++.h>

using namespace std;
#define int long long
#define rep(i,n) REP(i,0,n)
#define REP(i,s,e) for(int i=(s); i<(int)(e); i++)
#define repr(i, n) REPR(i, n, 0)
#define REPR(i, s, e) for(int i=(int)(s-1); i>=(int)(e); i--)
#define pb push_back
#define all(r) r.begin(),r.end()
#define rall(r) r.rbegin(),r.rend()
#define fi first
#define se second

typedef long long ll;
typedef vector<int> vi;
typedef vector<ll> vl;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;

const int INF = 1e9;
const ll MOD = 1e9 + 7;
double EPS = 1e-8;

const int sqrtN = 350;
const int MAX_N = sqrtN * sqrtN + 1;
using TYPE = ll;
const TYPE DFLT = -1e18;    // max => -INF, min => INF, sum => 0

// change f1, f2
inline void f1(TYPE& a, const TYPE& b) {
    a+=b;
}
inline void f2(TYPE& a, const TYPE& b) {
    a = max(a, b);
}

TYPE dat[MAX_N];
TYPE ans[sqrtN];
TYPE lazy[sqrtN];

inline int get_i(int x) {return x / sqrtN;}
inline int get_j(int x) {return x % sqrtN;}
inline int get_idx(int i, int j) {return i * sqrtN + j;}


void calcLazy(int i) {
    REP(j, 0, sqrtN) {
        f1(dat[get_idx(i, j)], lazy[i]);
    }
    lazy[i] = 0;
}

void add(int l, int r, TYPE s) {
    REP(i, get_i(l), sqrtN) {
        int x = get_idx(i, 0), y = x + sqrtN;
        if (r <= x) break;
        if (l <= x && y <= r) {
            f1(ans[i], s);
            f1(lazy[i], s);
        } else {
            calcLazy(i);
            REP(idx, max(x, l), min(y, r)) {
                f1(dat[idx], s);
            }
        }
    }
}

TYPE query(int l, int r) {
    TYPE ret = DFLT;
    REP(i, get_i(l), sqrtN) {
        int x = get_idx(i, 0), y = x + sqrtN;
        if (r <= x) break;
        if (l <= x && y <= r) {
            f2(ret, ans[i]);
        } else {
            calcLazy(i);
            REP(idx, max(x, l), min(y, r)) {
                f2(ret, dat[idx]);
            }
        }
    }
    return ret;
}

signed main() {
    int n;
    cin >> n;
    rep(i, n - 1) {
        ll a;
        cin >> a;
        dat[i] = a + 3 * (n - i - 1);
    }
    rep(i, n) ans[get_i(i)] = max(ans[get_i(i)], dat[i]);
    int m;
    cin >> m;
    rep(i, m) {
        int l, r, d;
        cin >> l >> r >> d;
        --l;
        add(l, r, (ll)d);
        cout << query(0, n) << endl;
    }
    return 0;
}