Greg has an array a = a1, a2, ..., an and m operations. Each operation looks as: li, ri, di, (1 ≤ li ≤ ri ≤ n). To apply operation i to the array means to increase all array elements with numbers li, li + 1, ..., ri by value di.

Greg wrote down k queries on a piece of paper. Each query has the following form: xi, yi, (1 ≤ xi ≤ yi ≤ m). That means that one should apply operations with numbers xi, xi + 1, ..., yi to the array.

Now Greg is wondering, what the array a will be after all the queries are executed. Help Greg.

Input

The first line contains integers n, m, k (1 ≤ n, m, k ≤ 105). The second line contains n integers: a1, a2, ..., an (0 ≤ ai ≤ 105) — the initial array.

Next m lines contain operations, the operation number i is written as three integers: li, ri, di, (1 ≤ li ≤ ri ≤ n), (0 ≤ di ≤ 105).

Next k lines contain the queries, the query number i is written as two integers: xi, yi, (1 ≤ xi ≤ yi ≤ m).

The numbers in the lines are separated by single spaces.

Output

On a single line print n integers a1, a2, ..., an — the array after executing all the queries. Separate the printed numbers by spaces.

Please, do not use the %lld specifier to read or write 64-bit integers in C++. It is preferred to use the cin, cout streams of the %I64d specifier.

Examples

Input

3 3 3
1 2 3
1 2 1
1 3 2
2 3 4
1 2
1 3
2 3


Output

9 18 17


Input

1 1 1
1
1 1 1
1 1


Output

2


Input

4 3 6
1 2 3 4
1 2 1
2 3 2
3 4 4
1 2
1 3
2 3
1 2
1 3
2 3


Output

5 18 31 20
#include <bits/stdc++.h>
using namespace std;
vector<long long> tree;
void build(vector<long long> &arr, long long ind, long long l, long long r) {
  if (l == r) {
    tree[ind] = arr[l];
    return;
  }
  long long mid = (l + r) / 2;
  build(arr, ind * 2, l, mid);
  build(arr, ind * 2 + 1, mid + 1, r);
  tree[ind] = max(tree[ind * 2], tree[ind * 2 + 1]);
}
long long getMax(long long ind, long long tl, long long tr, long long l,
                 long long r) {
  if (l > r) return 0;
  if (l == tl && r == tr) return tree[ind];
  long long tm = (tl + tr) / 2;
  long long a = getMax(ind * 2, tl, tm, l, min(r, tm));
  long long b = getMax(ind * 2 + 1, tm + 1, tr, max(l, tm + 1), r);
  return max(a, b);
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  long long n, m, k;
  cin >> n >> m >> k;
  vector<long long> arr(n);
  for (int i = 0; i < n; i++) {
    cin >> arr[i];
  }
  vector<pair<pair<long long, long long>, long long>> oper;
  for (int i = 0; i < m; i++) {
    long long l, r, d;
    cin >> l >> r >> d;
    l--;
    r--;
    oper.push_back({{l, r}, d});
  }
  vector<pair<long long, long long>> zapr(m, {0, 0});
  for (int i = 0; i < k; i++) {
    long long x, y;
    cin >> x >> y;
    x--;
    y--;
    zapr[x].first++;
    zapr[y].second++;
  }
  long long sum = 0;
  vector<long long> count(m);
  for (int i = 0; i < m; i++) {
    sum += zapr[i].first;
    count[i] = sum;
    sum -= zapr[i].second;
  }
  vector<pair<long long, long long>> zadd(n, {0, 0});
  for (int i = 0; i < m; i++) {
    auto p = oper[i];
    long long d = p.second * count[i];
    long long l = p.first.first;
    long long r = p.first.second;
    zadd[l].first += d;
    zadd[r].second += d;
  }
  sum = 0;
  for (int i = 0; i < n; i++) {
    sum += zadd[i].first;
    arr[i] += sum;
    sum -= zadd[i].second;
  }
  for (int i = 0; i < n; i++) {
    cout << arr[i] << " ";
  }
  return 0;
}
