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>
int tm[10][100008], a[100008];
long long ans[100008], t[100008];
int main() {
  int n, m, op;
  while (~scanf("%d%d%d", &n, &m, &op)) {
    for (int i = 1; i <= n; i++) {
      scanf("%d", &a[i]);
    }
    memset(ans, 0, sizeof(ans));
    memset(t, 0, sizeof(t));
    for (int i = 1; i <= m; i++) {
      scanf("%d%d%d", &tm[0][i], &tm[1][i], &tm[2][i]);
    }
    int x, y, z;
    while (op--) {
      scanf("%d%d", &x, &y);
      ans[y + 1]--;
      ans[x]++;
    }
    for (int i = 1; i <= m; i++) ans[i] += ans[i - 1];
    for (int i = 1; i <= m; i++) {
      x = tm[0][i];
      y = tm[1][i];
      z = tm[2][i];
      long long pk = 1;
      long long qq = (long long)z * ans[i];
      t[y + 1] -= qq;
      t[x] += qq;
    }
    for (int i = 1; i <= n; i++) t[i] += t[i - 1];
    for (int i = 1; i <= n; i++) printf("%lld ", t[i] + a[i]);
    printf("\n");
  }
}
