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;
const int maxN = (int)1E5 + 5;
const int maxM = (int)1E5 + 5;
int n, m, k;
int a[maxN];
long long dd[maxN];
pair<pair<int, int>, int> b[maxM];
int d[maxM];
int main() {
  scanf("%d %d %d", &n, &m, &k);
  for (int i = 1; i <= n; ++i) scanf("%d", &a[i]);
  for (int i = 1; i <= m; ++i)
    scanf("%d %d %d", &b[i].first.first, &b[i].first.second, &b[i].second);
  memset(d, 0, sizeof(d));
  for (int i = 1; i <= k; ++i) {
    int x, y;
    scanf("%d %d", &x, &y);
    ++d[x];
    --d[y + 1];
  }
  int s = 0;
  for (int i = 1; i <= m; ++i) {
    s += d[i];
    d[i] = s;
  }
  memset(dd, 0, sizeof(dd));
  for (int i = 1; i <= m; ++i) {
    dd[b[i].first.first] += (long long)d[i] * b[i].second;
    dd[b[i].first.second + 1] -= (long long)d[i] * b[i].second;
  }
  long long ss = 0;
  for (int i = 1; i <= n; ++i) {
    ss += dd[i];
    cout << ss + a[i] << " ";
  }
  cout << "\n";
  return 0;
}
