The Cartesian coordinate system is set in the sky. There you can see n stars, the i-th has coordinates (xi, yi), a maximum brightness c, equal for all stars, and an initial brightness si (0 ≤ si ≤ c).

Over time the stars twinkle. At moment 0 the i-th star has brightness si. Let at moment t some star has brightness x. Then at moment (t + 1) this star will have brightness x + 1, if x + 1 ≤ c, and 0, otherwise.

You want to look at the sky q times. In the i-th time you will look at the moment ti and you will see a rectangle with sides parallel to the coordinate axes, the lower left corner has coordinates (x1i, y1i) and the upper right — (x2i, y2i). For each view, you want to know the total brightness of the stars lying in the viewed rectangle.

A star lies in a rectangle if it lies on its border or lies strictly inside it.

Input

The first line contains three integers n, q, c (1 ≤ n, q ≤ 105, 1 ≤ c ≤ 10) — the number of the stars, the number of the views and the maximum brightness of the stars.

The next n lines contain the stars description. The i-th from these lines contains three integers xi, yi, si (1 ≤ xi, yi ≤ 100, 0 ≤ si ≤ c ≤ 10) — the coordinates of i-th star and its initial brightness.

The next q lines contain the views description. The i-th from these lines contains five integers ti, x1i, y1i, x2i, y2i (0 ≤ ti ≤ 109, 1 ≤ x1i < x2i ≤ 100, 1 ≤ y1i < y2i ≤ 100) — the moment of the i-th view and the coordinates of the viewed rectangle.

Output

For each view print the total brightness of the viewed stars.

Examples

Input

2 3 3
1 1 1
3 2 0
2 1 1 2 2
0 2 1 4 5
5 1 1 5 5


Output

3
0
3


Input

3 4 5
1 1 2
2 3 0
3 3 1
0 1 1 100 100
1 2 2 4 4
2 2 1 4 7
1 50 50 51 51


Output

3
3
5
0

Note

Let's consider the first example.

At the first view, you can see only the first star. At moment 2 its brightness is 3, so the answer is 3.

At the second view, you can see only the second star. At moment 0 its brightness is 0, so the answer is 0.

At the third view, you can see both stars. At moment 5 brightness of the first is 2, and brightness of the second is 1, so the answer is 3.
#include <bits/stdc++.h>
using namespace std;
template <class T>
ostream &operator<<(ostream &os, vector<T> V) {
  os << "[";
  for (auto const &vv : V) os << vv << ",";
  os << "]";
  return os;
}
template <class L, class R>
ostream &operator<<(ostream &os, pair<L, R> P) {
  os << "(" << P.first << "," << P.second << ")";
  return os;
}
template <typename T>
inline void print_128(T num) {
  if (!num) return;
  print_128(num / 10);
  cout.put((num % 10) + '0');
}
inline int fstoi(const string &str) {
  auto it = str.begin();
  bool neg = 0;
  int num = 0;
  if (*it == '-')
    neg = 1;
  else
    num = *it - '0';
  ++it;
  while (it < str.end()) num = num * 10 + (*it++ - '0');
  if (neg) num *= -1;
  return num;
}
inline void getch(char &x) {
  while (x = getchar_unlocked(), x < 33) {
    ;
  }
}
inline void getstr(string &str) {
  str.clear();
  char cur;
  while (cur = getchar_unlocked(), cur < 33) {
    ;
  }
  while (cur > 32) {
    str += cur;
    cur = getchar_unlocked();
  }
}
template <typename... Args>
inline void getstr(string &str, Args &...arg) {
  str.clear();
  char cur;
  while (cur = getchar_unlocked(), cur < 33) {
    ;
  }
  while (cur > 32) {
    str += cur;
    cur = getchar_unlocked();
  }
  getstr(arg...);
}
template <typename T>
inline bool sc(T &num) {
  bool neg = 0;
  int c;
  num = 0;
  while (c = getchar_unlocked(), c < 33) {
    if (c == EOF) return false;
  }
  if (c == '-') {
    neg = 1;
    c = getchar_unlocked();
  }
  for (; c > 47; c = getchar_unlocked()) num = num * 10 + c - 48;
  if (neg) num *= -1;
  return true;
}
template <typename T, typename... Args>
inline void sc(T &num, Args &...args) {
  bool neg = 0;
  int c;
  num = 0;
  while (c = getchar_unlocked(), c < 33) {
    ;
  }
  if (c == '-') {
    neg = 1;
    c = getchar_unlocked();
  }
  for (; c > 47; c = getchar_unlocked()) num = num * 10 + c - 48;
  if (neg) num *= -1;
  sc(args...);
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int n, q, c;
  sc(n, q, c);
  int rem = c + 1;
  static int res[12][101][101];
  int x, y, s;
  while (n--) {
    sc(x, y, s);
    int i = 0;
    while (i < rem) {
      res[i++][y][x] += (s++) % rem;
    }
  }
  for (auto &i : res) {
    for (auto &x : i) {
      for (int i = (1); i < int((101)); ++i) {
        x[i] += x[i - 1];
      }
    }
  }
  int x1, x2, y1, y2, t;
  while (q--) {
    sc(t, x1, y1, x2, y2);
    int sum = 0;
    int actt = t % rem;
    while (y1 <= y2) {
      if (x1)
        sum += res[actt][y1][x2] - res[actt][y1][x1 - 1];
      else
        sum += res[actt][y1][x2];
      ++y1;
    }
    cout << sum << '\n';
  }
}
