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;
struct op {
  long long l, r, d;
};
long long v[112345], aux[112345], auxl[112345];
int main() {
  long long n, m, k, l, r, d, i, x, y, sa = 0;
  op a[112345];
  scanf("%I64d", &n);
  scanf("%I64d", &m);
  scanf("%I64d", &k);
  for (i = 1; i < (n + 1); i++) {
    scanf("%I64d", &v[i]);
  }
  for (i = 1; i < (m + 1); i++) {
    scanf("%I64d", &a[i].l);
    scanf("%I64d", &a[i].r);
    scanf("%I64d", &a[i].d);
  }
  for (i = 0; i < k; i++) {
    scanf("%I64d", &x);
    scanf("%I64d", &y);
    aux[x] += 1;
    aux[y + 1] -= 1;
  }
  for (i = 1; i < (m + 1); i++) {
    sa += aux[i];
    a[i].d *= sa;
  }
  for (i = 1; i < (m + 1); i++) {
    auxl[a[i].l] += a[i].d;
    auxl[a[i].r + 1] -= a[i].d;
  }
  sa = 0;
  for (i = 1; i < (n + 1); i++) {
    sa += auxl[i];
    v[i] += sa;
    printf("%I64d ", v[i]);
  }
  printf("\n");
  return (0);
}
