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.*;
import java.math.*;
public class Class{
    public static void main(String[] args) throws IOException{
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        StringTokenizer st = new StringTokenizer(br.readLine());
        int n = Integer.parseInt(st.nextToken());
        int m = Integer.parseInt(st.nextToken());
        int k = Integer.parseInt(st.nextToken());
        st = new StringTokenizer(br.readLine());
        long[] a = new long[n+2];
        for(int i=1; i<=n; i++)a[i] = Long.parseLong(st.nextToken());
        String[] str = new String[m+1];
        for(int i=1; i<=m; i++)str[i]=br.readLine();
        long[] count = new long[m+2];
        for(int i=1; i<=k; i++){
            st = new StringTokenizer(br.readLine());
            int x = Integer.parseInt(st.nextToken());
            int y = Integer.parseInt(st.nextToken());
            count[x]++;
            count[y+1]--;
        }
        for(int i=1; i<m+2; i++)count[i]+=count[i-1];
        long[] add = new long[n+2];
        for(int i=1; i<=m; i++){
            st = new StringTokenizer(str[i]);
            int l = Integer.parseInt(st.nextToken());
            int r = Integer.parseInt(st.nextToken());
            int d = Integer.parseInt(st.nextToken());
            add[l] += count[i]*d;
            add[r+1] -= count[i]*d;
        }
        for(int i=1; i<=n; i++)add[i]+=add[i-1];
        StringBuffer sb = new StringBuffer();
        for(int i=1; i<=n; i++){a[i]+=add[i]; sb.append(a[i]).append(" ");}
        System.out.print(sb);        
    }
    
    static class Pair{
        long start;
        long end;
    
        public Pair(long start, long end){
            this.start = start;
            this.end = end;
        }
        
        public int hashCode() {
            return (int)(this.start + this.end);
        }
        
        public boolean equals(Object p){
            if(((Pair)p).start==this.start && ((Pair)p).end==this.end)return true;
            return false;
        }
        
        public String toString(){
            return this.start+" "+this.end;
        }
    }
}