#include <bits/stdc++.h>
using namespace std;
typedef struct op {
  long long int l;
  long long int r;
  long long int d;
} op;
op ops[100005];
long long int v[100005], inc[100005], inc2[100005], rp[100005], c, n, k, m, l,
    r, ans[100005];
int main() {
  scanf("%I64d %I64d %I64d", &n, &m, &k);
  for (long long int i = 1; i <= n; i++) {
    scanf("%I64d", &v[i]);
  }
  for (long long int i = 1; i <= m; i++) {
    scanf("%I64d %I64d %I64d", &ops[i].l, &ops[i].r, &ops[i].d);
  }
  for (long long int i = 1; i <= k; i++) {
    scanf("%I64d %I64d", &l, &r);
    inc[l] += 1;
    inc[r + 1] -= 1;
  }
  for (long long int i = 1; i <= m; i++) {
    c += inc[i];
    rp[i] += c;
  }
  for (long long int i = 1; i <= m; i++) {
    inc2[ops[i].l] += rp[i] * ops[i].d;
    inc2[ops[i].r + 1] -= rp[i] * ops[i].d;
  }
  c = 0;
  for (long long int i = 1; i <= n; i++) {
    c += inc2[i];
    ans[i] = c + v[i];
  }
  for (long long int i = 1; i <= n; i++) {
    printf("%I64d ", ans[i]);
  }
  printf("\n");
  return 0;
}

### Prompt

Create a solution in Cpp for 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;
typedef struct op {
  long long int l;
  long long int r;
  long long int d;
} op;
op ops[100005];
long long int v[100005], inc[100005], inc2[100005], rp[100005], c, n, k, m, l,
    r, ans[100005];
int main() {
  scanf("%I64d %I64d %I64d", &n, &m, &k);
  for (long long int i = 1; i <= n; i++) {
    scanf("%I64d", &v[i]);
  }
  for (long long int i = 1; i <= m; i++) {
    scanf("%I64d %I64d %I64d", &ops[i].l, &ops[i].r, &ops[i].d);
  }
  for (long long int i = 1; i <= k; i++) {
    scanf("%I64d %I64d", &l, &r);
    inc[l] += 1;
    inc[r + 1] -= 1;
  }
  for (long long int i = 1; i <= m; i++) {
    c += inc[i];
    rp[i] += c;
  }
  for (long long int i = 1; i <= m; i++) {
    inc2[ops[i].l] += rp[i] * ops[i].d;
    inc2[ops[i].r + 1] -= rp[i] * ops[i].d;
  }
  c = 0;
  for (long long int i = 1; i <= n; i++) {
    c += inc2[i];
    ans[i] = c + v[i];
  }
  for (long long int i = 1; i <= n; i++) {
    printf("%I64d ", ans[i]);
  }
  printf("\n");
  return 0;
}

```