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.
def fun(n,q,c,stars,querry):
    m=100
    precom=[[[0]*m for _ in range(m)] for _ in range(c+1)]
    for x,y,p in stars:
        for i in range(c+1):
            temp=(p+i)%(c+1)
            precom[i][x-1][y-1]+=temp
    for k in range(c+1):
        for i in range(0,m):
            for j in range(1,m):
                precom[k][i][j]+=precom[k][i][j-1]
    
    for k in range(c+1):
        for j in range(0,m):
            for i in range(1,m):
                precom[k][i][j]+= precom[k][i-1][j]
    
    for t,x1,y1,x2,y2 in querry:
        t=t%(c+1)
        x1,y1,x2,y2,a,b,C,d=x1-1,y1-1,x2-1,y2-1,0,0,0,0
        a=precom[t][x2][y2]
        if x1!=0:
            b=precom[t][x1-1][y2]
        if y1!=0:
            C=precom[t][x2][y1-1]
        if x1!=0 and  y1!=0:
            d=precom[t][x1-1][y1-1]
        print(a-b-C+d)
    
n,q,c=list(map(lambda x:int(x),input().split()))
stars=[list(map(lambda x:int(x),input().split())) for _ in range(n)]
querry=[list(map(lambda x:int(x),input().split())) for _ in range(q)]
fun(n,q,c,stars,querry)