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
import java.util.*;
import java.io.*;
public class a {
public static void main(String[] args) throws IOException
{
    //Scanner input = new Scanner(new File("input.txt"));
    //PrintWriter out = new PrintWriter(new File("output.txt"));
  input.init(System.in);
  PrintWriter out = new PrintWriter(System.out);
  
  int n= input.nextInt(), m = input.nextInt(), k = input.nextInt();
  long[] data = new long[n];
  long[] diffs = new long[n];
  for(int i = 0; i<n; i++)
      data[i] = input.nextInt();
  int[] ls = new int[m], rs = new int[m], ds = new int[m];
  for(int i = 0; i<m; i++)
  {
      ls[i] = input.nextInt()-1;
      rs[i] = input.nextInt()-1;
      ds[i] = input.nextInt();
  }
  int[] qcount = new int[m];
  for(int i = 0; i<k; i++)
  {
      int a = input.nextInt()-1, b = input.nextInt()-1;
      qcount[a] ++;
      if(b<m-1) qcount[b+1]--;
  }
  long x = 0;
  for(int i = 0; i<m; i++)
  {
      x += qcount[i];
      diffs[ls[i]] += ds[i]*x;
      if(rs[i]<n-1) diffs[rs[i]+1] -= ds[i]*x;
  }
  x = 0;
  for(int i = 0; i<n; i++)
  {
      x += diffs[i];
      data[i] += x;
  }
  for(int i = 0; i<n; i++) out.print(data[i]+" ");
  
  out.close();
  
}

static class input {
  static BufferedReader reader;
  static StringTokenizer tokenizer;

  /** call this method to initialize reader for InputStream */
  static void init(InputStream input) {
      reader = new BufferedReader(
                   new InputStreamReader(input) );
      tokenizer = new StringTokenizer("");
  }

  /** get next word */
  static String next() throws IOException {
      while ( ! tokenizer.hasMoreTokens() ) {
          //TODO add check for eof if necessary
          tokenizer = new StringTokenizer(
                 reader.readLine() );
      }
      return tokenizer.nextToken();
  }

  static int nextInt() throws IOException {
      return Integer.parseInt( next() );
  }
  
  static double nextDouble() throws IOException {
      return Double.parseDouble( next() );
  }
  static long nextLong() throws IOException {
      return Long.parseLong( next() );
  }
  static String nextLine() throws IOException {
    return reader.readLine();
  }
}
}