力竭，97分极限。

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int MAXN = 300005;

int n, m, k;
int owner[MAXN];
ll need[MAXN];
int ans[MAXN];
vector<int> stations[MAXN];

struct Event {
    int l, r;
    ll a;
} rain[MAXN];

ll bit[MAXN];
void add(int x, ll v) {
    for (; x <= m; x += x & -x) bit[x] += v;
}
ll query(int x) {
    ll res = 0;
    for (; x; x -= x & -x) res += bit[x];
    return res;
}
void apply(int id, int coeff) {
    int l = rain[id].l, r = rain[id].r;
    ll a = rain[id].a * coeff;
    if (l <= r) {
        add(l, a);
        if (r + 1 <= m) add(r + 1, -a);
    } else {
        add(1, a);
        if (r + 1 <= m) add(r + 1, -a);
        add(l, a);
    }
}

void solve(int l, int r, vector<int>& ids) {
    if (l > r || ids.empty()) return;
    if (l == r) {
        for (int id : ids) ans[id] = l;
        return;
    }
    int mid = (l + r) >> 1;
    // 将[l, mid]的陨石雨加入
    for (int i = l; i <= mid; ++i) apply(i, 1);
    
    vector<int> leftIds, rightIds;
    for (int id : ids) {
        ll sum = 0;
        for (int pos : stations[id]) {
            sum += query(pos);
            if (sum >= need[id]) break;
        }
        if (sum >= need[id]) {
            leftIds.push_back(id);
        } else {
            rightIds.push_back(id);
        }
    }
    // 撤销[l, mid]的陨石雨
    for (int i = l; i <= mid; ++i) apply(i, -1);
    
    solve(l, mid, leftIds);
    // 注意：递归右半边时，需要将[l, mid]的陨石雨再次加入
    for (int i = l; i <= mid; ++i) apply(i, 1);
    solve(mid + 1, r, rightIds);
    // 撤销[l, mid]的陨石雨，恢复状态
    for (int i = l; i <= mid; ++i) apply(i, -1);
}

int main() {
    scanf("%d%d", &n, &m);
    for (int i = 1; i <= m; ++i) {
        int o;
        scanf("%d", &o);
        stations[o].push_back(i);
    }
    for (int i = 1; i <= n; ++i) {
        scanf("%lld", &need[i]);
    }
    scanf("%d", &k);
    for (int i = 1; i <= k; ++i) {
        scanf("%d%d%lld", &rain[i].l, &rain[i].r, &rain[i].a);
    }
    // 添加一场虚拟陨石雨，用于处理无解
    rain[k + 1] = {1, 1, 0};  // 注意：这里可以任意设置，因为雨量为0
    ++k;
    
    vector<int> allIds;
    for (int i = 1; i <= n; ++i) {
        allIds.push_back(i);
        ans[i] = k + 1;  // 初始化为k+1表示无解
    }
    
    solve(1, k, allIds);
    
    for (int i = 1; i <= n; ++i) {
        if (ans[i] == k) {  // 注意：k是虚拟陨石雨的编号
            printf("NIE\n");
        } else {
            printf("%d\n", ans[i]);
        }
    }
    return 0;
}