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;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  long long N, M, K;
  scanf("%lld", &N);
  scanf("%lld", &M);
  scanf("%lld", &K);
  long long Inp[M + 5][3], temp[100005] = {0};
  vector<long long> a, b;
  for (long long n = 0; n < N; n = n + 1) {
    long long x;
    scanf("%lld", &x);
    a.push_back(x);
  }
  for (long long n = 0; n < M; n = n + 1) {
    scanf("%lld", &Inp[n][0]);
    scanf("%lld", &Inp[n][1]);
    scanf("%lld", &Inp[n][2]);
    Inp[n][0] -= 1;
    Inp[n][1] -= 1;
  }
  for (long long n = 0; n < K; n = n + 1) {
    long long x, y;
    scanf("%lld", &x);
    scanf("%lld", &y);
    x -= 1;
    y -= 1;
    temp[x] += 1;
    if ((y + 1) < M) {
      temp[y + 1] -= 1;
    }
  }
  for (long long n = 0; n < M; n = n + 1) {
    if (n == 0) {
      b.push_back(temp[n]);
    } else {
      b.push_back(b[b.size() - 1] + temp[n]);
    }
  }
  long long Sum[100005] = {0};
  for (long long n = 0; n < M; n = n + 1) {
    long long x, y, z;
    x = Inp[n][0];
    y = Inp[n][1];
    z = Inp[n][2];
    Sum[x] += (z * b[n]);
    if ((y + 1) < N) {
      Sum[y + 1] -= (z * b[n]);
    }
  }
  vector<long long> c;
  for (long long n = 0; n < N; n = n + 1) {
    if (n == 0) {
      c.push_back(Sum[n]);
    } else {
      c.push_back(c[c.size() - 1] + Sum[n]);
    }
  }
  for (long long n = 0; n < N; n = n + 1) {
    cout << a[n] + c[n] << " ";
  }
  cout << endl;
  return 0;
}
