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 x[100005];
long long q[100005];
long long u[100005];
int main() {
  long long n, m, k;
  cin >> n >> m >> k;
  memset(q, 0, sizeof(q));
  memset(u, 0, sizeof(u));
  for (long long i = 0; i < n; i++) {
    cin >> x[i];
  }
  vector<long long> b, c, d;
  for (long long i = 0; i < m; i++) {
    long long w, e, r;
    cin >> w >> e >> r;
    w--;
    e--;
    b.push_back(w);
    c.push_back(e);
    d.push_back(r);
  }
  for (long long i = 0; i < k; i++) {
    long long o, p;
    cin >> o >> p;
    o--;
    p--;
    q[o]++;
    q[p + 1]--;
  }
  long long sum = 0;
  for (long long i = 0; i < m; i++) {
    sum += q[i];
    q[i] = sum;
  }
  for (long long i = 0; i < m; i++) {
    long long v = q[i];
    u[b[i]] = u[b[i]] + v * (d[i]);
    u[c[i] + 1] = u[c[i] + 1] - (v * d[i]);
  }
  long long sum1 = 0;
  for (long long i = 0; i < n; i++) {
    sum1 += u[i];
    u[i] = sum1;
  }
  for (long long i = 0; i < n; i++) {
    x[i] = x[i] + u[i];
  }
  for (long long i = 0; i < n; i++) {
    cout << x[i] << " ";
  }
  cout << endl;
}
