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>
namespace fastIO {
bool IOerror = 0;
inline char nc() {
  static char buf[100000], *p1 = buf + 100000, *pend = buf + 100000;
  if (p1 == pend) {
    p1 = buf;
    pend = buf + fread(buf, 1, 100000, stdin);
    if (pend == p1) {
      IOerror = 1;
      return -1;
    }
  }
  return *p1++;
}
inline bool blank(char ch) {
  return ch == ' ' || ch == '\n' || ch == '\r' || ch == '\t';
}
template <class T>
inline bool read(T& x) {
  bool sign = 0;
  char ch = nc();
  x = 0;
  for (; blank(ch); ch = nc())
    ;
  if (IOerror) return false;
  if (ch == '-') sign = 1, ch = nc();
  for (; ch >= '0' && ch <= '9'; ch = nc()) x = x * 10 + ch - '0';
  if (sign) x = -x;
  return true;
}
inline bool read(double& x) {
  bool sign = 0;
  char ch = nc();
  x = 0;
  for (; blank(ch); ch = nc())
    ;
  if (IOerror) return false;
  if (ch == '-') sign = 1, ch = nc();
  for (; ch >= '0' && ch <= '9'; ch = nc()) x = x * 10 + ch - '0';
  if (ch == '.') {
    double tmp = 1;
    ch = nc();
    for (; ch >= '0' && ch <= '9'; ch = nc())
      tmp /= 10.0, x += tmp * (ch - '0');
  }
  if (sign) x = -x;
  return true;
}
inline bool read(char* s) {
  char ch = nc();
  for (; blank(ch); ch = nc())
    ;
  if (IOerror) return false;
  for (; !blank(ch) && !IOerror; ch = nc()) *s++ = ch;
  *s = 0;
  return true;
}
inline bool read(char& c) {
  for (c = nc(); blank(c); c = nc())
    ;
  if (IOerror) {
    c = -1;
    return false;
  }
  return true;
}
template <class T, class... U>
bool read(T& h, U&... t) {
  return read(h) && read(t...);
}
};  // namespace fastIO
using namespace fastIO;
using namespace std;
const long long N = 1e2 + 5;
const double eps = 1e-7;
const double pi = acos(-1.0);
const long long mod = 998244353;
long long sum[N][N][15];
long long n, q, c;
void f() {
  for (long long x = 1; x <= 100; x++)
    for (long long y = 1; y <= 100; y++)
      for (long long c = 0; c <= 10; c++)
        sum[x][y][c] +=
            sum[x - 1][y][c] + sum[x][y - 1][c] - sum[x - 1][y - 1][c];
}
signed main() {
  read(n, q, c);
  for (long long i = 1; i <= n; i++) {
    long long ini, x, y;
    read(x, y, ini);
    sum[x][y][ini]++;
  }
  f();
  while (q--) {
    long long t, x1, y1, x2, y2, res = 0;
    read(t, x1, y1, x2, y2);
    for (long long tc = 0; tc <= c; tc++) {
      long long nc = (tc + t) % (c + 1);
      res += nc * (sum[x2][y2][tc] - sum[x2][y1 - 1][tc] - sum[x1 - 1][y2][tc] +
                   sum[x1 - 1][y1 - 1][tc]);
    }
    printf("%lld\n", res);
  }
  return 0;
}
