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.
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 ######   #########      #          #       #   #   #   #           #########    #   
 ######   #########     #           #       #    #  #   #           #########    #   
 #     #  #       #    #            #       #     # #   #    ####   #       #    #   
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from __future__ import print_function # for PyPy2
from itertools import permutations as perm
# from fractions import Fraction
from collections import *
from sys import stdin
from bisect import *
from heapq import *
from math import *
 
g   = lambda : stdin.readline().strip()
gl  = lambda : g().split()
gil = lambda : [int(var) for var in gl()]
gfl = lambda : [float(var) for var in gl()]
gcl = lambda : list(g())
gbs = lambda : [int(var) for var in g()]
mod = int(1e9)+7
inf = float("inf")

n, q, c = gil()

f = [ [[0 for _ in range(101)] for _ in range(101)] for _ in range(c+1) ]

for _ in range(n):
	x, y, s = gil()
	f[s][x][y] += 1

# pre computation
for y in range(1, 101):
	for x in range(1, 101):
		for ci in range(c+1):
			f[ci][x][y] += f[ci][x][y-1] + f[ci][x-1][y] - f[ci][x-1][y-1] 

for _ in range(q):
		t, x1, y1, x2, y2 = gil()
		t %= (c+1)
		val = 0
		for ci in range(c+1):
			fi = f[ci][x2][y2] + f[ci][x1-1][y1-1] - f[ci][x2][y1-1] - f[ci][x1-1][y2] 
			val += fi*( (t+ci)%(c+1) )
		print(val)
