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 ll = long long;
using ull = unsigned long long;
using std::cin;
using std::cout;
void print() {}
template <typename T, typename... Args>
void print(T t, Args... args) {
  cout << t << ' ', print(args...);
}
template <typename... Args>
void println(Args... args) {
  print(args...), cout << '\n';
}
void printF(const char *format) { cout << format; }
template <typename T, typename... Args>
void printF(const char *format, T t, Args... args) {
  while (*format != '%' && *format) cout.put(*format++);
  if (*format++ == '\0') return;
  cout << t;
  printF(format, args...);
}
ll getTotalBright(const std::vector<ll> &v, ll time, ll C) {
  ll ret = 0;
  for (ll i = 0; i < C + 1; ++i) {
    ret += v[i] * ((time + i) % (C + 1));
  }
  return ret;
}
int main() {
  std::ios_base::sync_with_stdio(false);
  std::cin.tie(NULL);
  int N, Q, C, x1, y1, x2, y2, s;
  int time;
  cin >> N >> Q >> C;
  std::array<std::array<std::vector<ll>, 101>, 101> grid;
  for (auto &row : grid)
    for (auto &col : row) col.resize(C + 1, 0);
  for (int i = 0; i < N; ++i) {
    cin >> x1 >> y1 >> s;
    grid[x1][y1][s]++;
  }
  for (int i = 1; i < 101; ++i) {
    for (int h = 0; h <= C; ++h) {
      grid[i][0][h] += grid[i - 1][0][h];
      grid[0][i][h] += grid[0][i - 1][h];
    }
  }
  for (int i = 1; i < 101; ++i) {
    for (int j = 1; j < 101; ++j) {
      for (int h = 0; h <= C; ++h) {
        grid[i][j][h] +=
            grid[i - 1][j][h] + grid[i][j - 1][h] - grid[i - 1][j - 1][h];
      }
    }
  }
  for (int q = 0; q < Q; ++q) {
    cin >> time >> x1 >> y1 >> x2 >> y2;
    ll bright = 0;
    bright += getTotalBright(grid[x2][y2], time, C);
    bright -= getTotalBright(grid[x1 - 1][y2], time, C);
    bright -= getTotalBright(grid[x2][y1 - 1], time, C);
    bright += getTotalBright(grid[x1 - 1][y1 - 1], time, C);
    println(bright);
  }
  return 0;
}
