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.
import random, math
from copy import deepcopy as dc

MAXX = 101
MAXY = 101


# Function to take input
def input_test():
	n, q, c = map(int, input().strip().split(" "))
	st = [[[0 for i in range(MAXX)] for j in range(MAXY)] for k in range(c+1)]
	
	# Taking Input and marking the values in the 
	# Frequency array and the grid that the star is present
	for _ in range(n):
		x, y, s = map(int, input().strip().split(" "))
		for t in range(c+1):
			st[t][x][y] += (s+t) % (c+1)

	for k in range(c + 1):
		for i in range(1,MAXX):
			for j in range(1,MAXY):
				st[k][i][j] = st[k][i][j] + st[k][i-1][j] + st[k][i][j-1] - st[k][i-1][j-1]
	

	comp = st
	# for i in comp:
	# 	for j in i:
	# 		print(j)
	# 	print("Done")

	# Taking the Queries
	for _ in range(q):
		t, x1, y1, x2, y2 = map(int, input().strip().split(" "))
		# t = t % (c+1)
		t = t % (c+1)
		to_be_add = comp[t][x1-1][y1-1]
		tot = comp[t][x2][y2] - comp[t][x2][y1-1] - comp[t][x1-1][y2] + to_be_add
		print(tot)




# Function to check test my code
def test():
	pass


input_test()
# test()