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;
long long BIT[111111];
int n = 0;
int BIT_Modify(int r, long long delta) {
  for (int i = r; i > 0; i -= (i & (-i))) BIT[i] += delta;
  return 0;
}
long long BIT_Query(int x) {
  if (x <= 0) return 0;
  long long sum = 0;
  for (int i = x; i <= n; i += (i & (-i))) sum += BIT[i];
  return sum;
}
long long BIT2[111111];
int m = 0;
int BIT2_Modify(int r, long long delta) {
  for (int i = r; i > 0; i -= (i & (-i))) BIT2[i] += delta;
  return 0;
}
long long BIT2_Query(int x) {
  if (x <= 0) return 0;
  long long sum = 0;
  for (int i = x; i <= m; i += (i & (-i))) sum += BIT2[i];
  return sum;
}
int l[111111];
int r[111111];
int d[111111];
int main(void) {
  int k = 0;
  scanf("%d %d %d", &n, &m, &k);
  for (int i = 1; i <= n; i++) {
    int x = 0;
    scanf("%d", &x);
    BIT_Modify(i, x);
    BIT_Modify(i - 1, -x);
  }
  for (int i = 1; i <= m; i++) {
    scanf("%d %d %d", &l[i], &r[i], &d[i]);
  }
  for (int i = 0; i < k; i++) {
    int x = 0;
    int y = 0;
    scanf("%d %d", &x, &y);
    BIT2_Modify(y, 1);
    BIT2_Modify(x - 1, -1);
  }
  for (int i = 1; i <= m; i++) {
    long long cnt = BIT2_Query(i);
    BIT_Modify(r[i], cnt * d[i]);
    BIT_Modify(l[i] - 1, -cnt * d[i]);
  }
  for (int i = 1; i <= n; i++) printf("%I64d ", BIT_Query(i));
  puts("");
  while (getchar() != EOF)
    ;
  return 0;
}
