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 math,sys,bisect,heapq
from collections import defaultdict,Counter,deque
from itertools import groupby,accumulate
#sys.setrecursionlimit(200000000)
int1 = lambda x: int(x) - 1
#input = iter(sys.stdin.buffer.read().decode().splitlines()).__next__
ilele = lambda: map(int,input().split())
alele = lambda: list(map(int, input().split()))
ilelec = lambda: map(int1,input().split())
alelec = lambda: list(map(int1, input().split()))
def list2d(a, b, c): return [[c] * b for i in range(a)]
def list3d(a, b, c, d): return [[[d] * c for j in range(b)] for i in range(a)]
#MOD = 1000000000 + 7
def Y(c):  print(["NO","YES"][c])
def y(c):  print(["no","yes"][c])
def Yy(c):  print(["No","Yes"][c])
    
dp = {}

def fun(A):
    B = list2d(101,101,0)
    for j in range(101):
        B[0][j] = A[0][j]
    for i in range(1,101):
        for j in range(101):
            B[i][j] = A[i][j] + B[i-1][j]
    for i in range(101):
        for j in range(1,101):
            B[i][j] = B[i][j] + B[i][j-1]
    return B

n,q,c = ilele()
G = defaultdict(list)
for i in range(n):
    a,b,s = ilele()
    G[(a,b)].append(s)
#print(G)
for i in range(c+1):
    #print(i,c)
    A = list2d(101,101,0)
    for j in range(1,101):
        for k in range(1,101):
            x  = G.get((j,k),[])
            for l in x:
                r = (l + i)%(c+1)
                #print(j,k,r,l,i)
                A[j][k] += r
    B = fun(A)
    dp[i] = B

for i in range(q):
    t,x1,y1,x2,y2 = ilele()
    t %= (c+1)
    #print(dp[t][x2][y2],dp[t][x2][y1])
    ans = dp[t][x2][y2]
    ans -= dp[t][x2][y1-1]
    ans -= dp[t][x1-1][y2]
    ans += dp[t][x1-1][y1-1]
    print(ans)