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;
const int inf = 1e9;
const long long int INF = 1e18;
const int N = 100005;
const int MOD = 1e9 + 7;
const double EPS = 1e-6;
const double PI = acos(-1.0);
long long int n, m, k;
long long int ara[N], add[N], Tree[4 * N], Lazy[4 * N];
struct query {
  int l, r;
  long long int val;
} Q[N];
void Propagate(int node, int b, int e) {
  if (Lazy[node] != 0) {
    Tree[node] += (Lazy[node] * (e - b + 1));
    if (b != e) {
      Lazy[2 * node] += Lazy[node];
      Lazy[2 * node + 1] += Lazy[node];
    }
    Lazy[node] = 0;
  }
}
void Update(int node, int b, int e, int i, int j, long long int val) {
  Propagate(node, b, e);
  if (i > e || j < b) return;
  if (i <= b && j >= e) {
    Tree[node] += val;
    if (b != e) {
      Lazy[2 * node] += val;
      Lazy[2 * node + 1] += val;
    }
    return;
  }
  int left = 2 * node, right = 2 * node + 1, mid = (b + e) / 2;
  Update(left, b, mid, i, j, val);
  Update(right, mid + 1, e, i, j, val);
  Tree[node] = Tree[left] + Tree[right];
}
long long int Query(int node, int b, int e, int pos) {
  Propagate(node, b, e);
  if (pos > e || pos < b) return 0;
  if (b == e && b == pos) return Tree[node];
  int left = 2 * node, right = 2 * node + 1, mid = (b + e) / 2;
  long long int q1 = Query(left, b, mid, pos);
  long long int q2 = Query(right, mid + 1, e, pos);
  return q1 + q2;
}
int main(int argc, char const *argv[]) {
  cin >> n >> m >> k;
  for (int i = 1; i <= n; i++) cin >> ara[i];
  for (int i = 1; i <= m; i++) {
    cin >> Q[i].l >> Q[i].r >> Q[i].val;
  }
  for (int i = 1; i <= k; i++) {
    int x, y;
    cin >> x >> y;
    add[y + 1]--, add[x]++;
  }
  for (int i = 1; i <= N; i++) {
    add[i] += add[i - 1];
  }
  for (int i = 1; i <= m; i++) {
    long long int tot = Q[i].val * add[i];
    Update(1, 1, n, Q[i].l, Q[i].r, tot);
  }
  for (int i = 1; i <= n; i++) {
    cout << Query(1, 1, n, i) + ara[i] << " ";
  }
  return 0;
}
