#include <bits/stdc++.h>
using namespace std;
int main() {
  int n, m, k;
  long long a[100001];
  long long p[100002];
  long long p2[100004];
  vector<tuple<int, int, int> > ops;
  scanf("%d %d %d", &n, &m, &k);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  for (int i = 0; i < m; i++) {
    int l, r, d;
    scanf("%d %d %d", &l, &r, &d);
    ops.push_back(make_tuple(l, r, d));
  }
  memset(p, 0, sizeof(p));
  memset(p2, 0, sizeof(p2));
  for (int i = 0; i < k; i++) {
    int x, y;
    cin >> x >> y;
    p2[x] += 1;
    p2[y + 1] -= 1;
  }
  for (int j = 1; j <= m; j++) {
    p2[j] += p2[j - 1];
    tuple<int, int, int> op = ops[(j - 1)];
    int l = get<0>(op), r = get<1>(op), d = get<2>(op);
    p[l] += (d * p2[j]);
    p[r + 1] -= (d * p2[j]);
  }
  for (int i = 1; i <= n; i++) {
    p[i] += p[i - 1];
    printf("%lld ", (a[i] + p[i]));
  }
  return 0;
}

### Prompt

Your challenge is to write a Cpp solution to the following problem:

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

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
  int n, m, k;
  long long a[100001];
  long long p[100002];
  long long p2[100004];
  vector<tuple<int, int, int> > ops;
  scanf("%d %d %d", &n, &m, &k);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  for (int i = 0; i < m; i++) {
    int l, r, d;
    scanf("%d %d %d", &l, &r, &d);
    ops.push_back(make_tuple(l, r, d));
  }
  memset(p, 0, sizeof(p));
  memset(p2, 0, sizeof(p2));
  for (int i = 0; i < k; i++) {
    int x, y;
    cin >> x >> y;
    p2[x] += 1;
    p2[y + 1] -= 1;
  }
  for (int j = 1; j <= m; j++) {
    p2[j] += p2[j - 1];
    tuple<int, int, int> op = ops[(j - 1)];
    int l = get<0>(op), r = get<1>(op), d = get<2>(op);
    p[l] += (d * p2[j]);
    p[r + 1] -= (d * p2[j]);
  }
  for (int i = 1; i <= n; i++) {
    p[i] += p[i - 1];
    printf("%lld ", (a[i] + p[i]));
  }
  return 0;
}

```