#include <bits/stdc++.h>
using namespace std;
int dx[] = {-1, 0, 0, 1};
int dy[] = {0, -1, 1, 0};
int dx1[] = {-1, -1, -1, 0, 0, 0, 1, 1, 1};
int dy1[] = {-1, 0, 1, -1, 0, 1, -1, 0, 1};
const double eps = 1e-9;
const int inf = 2000000000;
const long long int infLL = 9000000000000000000;
const long long int MOD = 1e+7;
const double PI = 3.141592653589793238462643383279;
int main() {
  long long int i, j, k;
  long long int n, m;
  long long int l, r;
  cin >> n >> m >> k;
  long long int a[n], lrd[m][3];
  long long int x[m + 1];
  memset(x, 0, sizeof(x));
  for (long long int i = 0; i < n; i++) cin >> a[i];
  for (long long int i = 0; i < m; i++)
    cin >> lrd[i][0] >> lrd[i][1] >> lrd[i][2];
  for (long long int i = 0; i < k; i++) {
    cin >> l >> r;
    l--;
    x[l]++;
    x[r]--;
  }
  for (long long int i = 1; i < m; i++) x[i] += x[i - 1];
  for (long long int i = 0; i < m; i++) lrd[i][2] *= x[i];
  long long int fin[n + 1];
  memset(fin, 0, sizeof(fin));
  for (long long int i = 0; i < m; i++) {
    fin[lrd[i][0] - 1] += lrd[i][2];
    fin[lrd[i][1]] -= lrd[i][2];
  }
  for (long long int i = 1; i < n; i++) fin[i] += fin[i - 1], a[i] += fin[i];
  a[0] += fin[0];
  for (long long int i = 0; i < n; i++) cout << a[i] << " ";
  return 0;
}

### Prompt

Please create a solution in Cpp 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 dx[] = {-1, 0, 0, 1};
int dy[] = {0, -1, 1, 0};
int dx1[] = {-1, -1, -1, 0, 0, 0, 1, 1, 1};
int dy1[] = {-1, 0, 1, -1, 0, 1, -1, 0, 1};
const double eps = 1e-9;
const int inf = 2000000000;
const long long int infLL = 9000000000000000000;
const long long int MOD = 1e+7;
const double PI = 3.141592653589793238462643383279;
int main() {
  long long int i, j, k;
  long long int n, m;
  long long int l, r;
  cin >> n >> m >> k;
  long long int a[n], lrd[m][3];
  long long int x[m + 1];
  memset(x, 0, sizeof(x));
  for (long long int i = 0; i < n; i++) cin >> a[i];
  for (long long int i = 0; i < m; i++)
    cin >> lrd[i][0] >> lrd[i][1] >> lrd[i][2];
  for (long long int i = 0; i < k; i++) {
    cin >> l >> r;
    l--;
    x[l]++;
    x[r]--;
  }
  for (long long int i = 1; i < m; i++) x[i] += x[i - 1];
  for (long long int i = 0; i < m; i++) lrd[i][2] *= x[i];
  long long int fin[n + 1];
  memset(fin, 0, sizeof(fin));
  for (long long int i = 0; i < m; i++) {
    fin[lrd[i][0] - 1] += lrd[i][2];
    fin[lrd[i][1]] -= lrd[i][2];
  }
  for (long long int i = 1; i < n; i++) fin[i] += fin[i - 1], a[i] += fin[i];
  a[0] += fin[0];
  for (long long int i = 0; i < n; i++) cout << a[i] << " ";
  return 0;
}

```