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;
using ll = long long;
using pii = pair<int, int>;
using pll = pair<long long, long long>;
using pcc = pair<char, char>;
using pdd = pair<double, double>;
using vi = vector<int>;
using vs = vector<string>;
using vll = vector<ll>;
using vpii = vector<pii>;
using vc = vector<char>;
using vd = vector<double>;
template <typename T>
T POW(T a, T b) {
  T x = 1;
  while (b) {
    if (b & 1) x *= a;
    a *= a;
    b >>= 1;
  }
  return x;
}
template <typename T>
T mPOW(T a, T b, T M) {
  T x = 1;
  while (b) {
    if (b & 1) x = (x * a) % M;
    a = (a * a) % M;
    b >>= 1;
  }
  return x;
}
template <typename T>
T Sqrt(T a) {
  T x = (T)sqrt(a);
  if ((x + 1) * (x + 1) == a) x++;
  return x;
}
int main(int argc, char *argv[]) {
  ll n, m, k;
  scanf("%lld%*c%lld%*c%lld%*c", &(n), &(m), &(k));
  vector<ll> ara(n + 5), opcnt(m + 5), add(n + 5);
  vector<pair<pll, ll>> op(m + 5);
  ll i, j;
  for (i = 1; i <= n; i++) {
    scanf("%lld%*c", &(ara[i]));
  }
  for (i = 1; i <= m; i++) {
    scanf("%lld%*c%lld%*c%lld%*c", &(op[i].first.first), &(op[i].first.second),
          &(op[i].second));
  }
  for (i = 1; i <= k; i++) {
    ll x, y;
    scanf("%lld%*c%lld%*c", &(x), &(y));
    opcnt[x]++;
    opcnt[y + 1]--;
  }
  for (i = 1; i <= m; i++) {
    opcnt[i] += opcnt[i - 1];
    add[op[i].first.first] += (opcnt[i] * op[i].second);
    add[op[i].first.second + 1] -= (opcnt[i] * op[i].second);
  }
  for (i = 1; i <= n; i++) {
    add[i] += add[i - 1];
    ara[i] += add[i];
    printf("%lld", (ara[i])), printf(" ");
  }
  return 0;
}
