#include <bits/stdc++.h>
using namespace std;
struct com {
  long long l, r, d;
};
int main() {
  long long n, i, j, m, k, x, y;
  static long long num_comands[110000], arr[110000], buf[110000];
  static com comands[110000];
  cin >> n >> m >> k;
  memset(arr, 0, sizeof(arr));
  for (i = 0; i < n; ++i) scanf("%I64d", &buf[i + 1]);
  for (i = 0; i < m; ++i)
    scanf("%I64d %I64d %I64d", &comands[i + 1].l, &comands[i + 1].r,
          &comands[i + 1].d);
  memset(num_comands, 0, sizeof(num_comands));
  for (i = 0; i < k; ++i) {
    scanf("%I64d %I64d", &x, &y);
    num_comands[x]++;
    num_comands[y + 1]--;
  }
  for (i = 1; i <= m; ++i) {
    num_comands[i] += num_comands[i - 1];
    comands[i].d *= num_comands[i];
  }
  for (i = 1; i <= m; ++i) {
    arr[comands[i].l] += comands[i].d;
    arr[comands[i].r + 1] -= comands[i].d;
  }
  for (i = 1; i <= n; ++i) {
    arr[i] += arr[i - 1];
    printf("%I64d ", arr[i] + buf[i]);
  }
  return 0;
}

### Prompt

Develop a solution in CPP to the problem described below:

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;
struct com {
  long long l, r, d;
};
int main() {
  long long n, i, j, m, k, x, y;
  static long long num_comands[110000], arr[110000], buf[110000];
  static com comands[110000];
  cin >> n >> m >> k;
  memset(arr, 0, sizeof(arr));
  for (i = 0; i < n; ++i) scanf("%I64d", &buf[i + 1]);
  for (i = 0; i < m; ++i)
    scanf("%I64d %I64d %I64d", &comands[i + 1].l, &comands[i + 1].r,
          &comands[i + 1].d);
  memset(num_comands, 0, sizeof(num_comands));
  for (i = 0; i < k; ++i) {
    scanf("%I64d %I64d", &x, &y);
    num_comands[x]++;
    num_comands[y + 1]--;
  }
  for (i = 1; i <= m; ++i) {
    num_comands[i] += num_comands[i - 1];
    comands[i].d *= num_comands[i];
  }
  for (i = 1; i <= m; ++i) {
    arr[comands[i].l] += comands[i].d;
    arr[comands[i].r + 1] -= comands[i].d;
  }
  for (i = 1; i <= n; ++i) {
    arr[i] += arr[i - 1];
    printf("%I64d ", arr[i] + buf[i]);
  }
  return 0;
}

```