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;
void func(void) {
  freopen("input.c", "r", stdin);
  freopen("output.c", "w", stdout);
}
const long long N = 1e5 + 10;
long long tree[N * 5][2], lazy[N * 5][2];
long long a[N], b[N], c[N];
void update(long long node, long long b, long long e, long long l, long long r,
            long long val, long long i) {
  if (lazy[node][i] != 0) {
    tree[node][i] += lazy[node][i];
    if (b != e) {
      lazy[node * 2][i] += lazy[node][i];
      lazy[node * 2 + 1][i] += lazy[node][i];
    }
    lazy[node][i] = 0;
  }
  if (b > e or b > r or e < l) return;
  if (b >= l and e <= r) {
    tree[node][i] += val;
    if (b != e) {
      lazy[node * 2][i] += val;
      lazy[node * 2 + 1][i] += val;
    }
    return;
  }
  long long left = node * 2;
  long long right = node * 2 + 1;
  long long mid = (b + e) / 2;
  update(left, b, mid, l, r, val, i), update(right, mid + 1, e, l, r, val, i);
  tree[node][i] = tree[left][i] + tree[right][i];
}
long long query(long long node, long long b, long long e, long long l,
                long long r, long long i) {
  if (lazy[node][i] != 0) {
    tree[node][i] += lazy[node][i];
    if (b != e) {
      lazy[node * 2][i] += lazy[node][i];
      lazy[node * 2 + 1][i] += lazy[node][i];
    }
    lazy[node][i] = 0;
  }
  if (b > e or b > r or e < l) return 0;
  if (b >= l and e <= r) {
    return tree[node][i];
  }
  long long left = node * 2;
  long long right = node * 2 + 1;
  long long mid = (b + e) / 2;
  return query(left, b, mid, l, r, i) + query(right, mid + 1, e, l, r, i);
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  long long n, q, i, j = 0, temp, t, k, ans = 0, sum = 0, x, y, z, cnt = 0, m,
                     fg = 0, mx = 0, mx1 = 0, mn = 8000000000000000000,
                     mn1 = 8000000000000000000;
  cin >> n >> m >> k;
  for (long long i = 0; i < n; i++) {
    cin >> x;
    update(1, 1, n, i + 1, i + 1, x, 0);
  }
  for (long long i = 1; i <= m; i++) {
    cin >> a[i] >> b[i] >> c[i];
  }
  for (long long i = 1; i <= k; i++) {
    cin >> x >> y;
    update(1, 1, m, x, y, 1, 1);
  }
  for (long long i = 1; i <= m; i++) {
    long long val = query(1, 1, m, i, i, 1);
    update(1, 1, n, a[i], b[i], c[i] * val, 0);
  }
  for (long long i = 1; i <= n; i++) {
    cout << query(1, 1, n, i, i, 0) << " ";
  }
}
