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 sys
input=sys.stdin.readline
n,q,c=map(int,input().split())
dp=[]
for i in range(101):
    k=[]
    for j in range(101):
        k.append([0]*(11))
    dp.append(k)
for i in range(n):
    x,y,s=map(int,input().split())
    dp[x][y][s]+=1
for i in range(1,101):
    for s in range(11):
        dp[i][0][s]+=dp[i-1][0][s]
for i in range(1,101):
    for s in range(11):
        dp[0][i][s]+=dp[0][i-1][s]
for j in range(101):
    for i in range(101):
        if i>0 and j>0:
            for s in range(11):
                dp[i][j][s]+=(dp[i-1][j][s]+dp[i][j-1][s]-dp[i-1][j-1][s])
for j in range(q):
    p,x1,y1,x2,y2=map(int,input().split())
    k=[0]*(11)
    ans=0
    for i in range(11):
        if x1>0 and y1>0:
            k[i]=(dp[x2][y2][i]-dp[x2][y1-1][i]-(dp[x1-1][y2][i]-dp[x1-1][y1-1][i]))
        elif x1>0 and y1==0:
            k[i] = (dp[x2][y2][i] - (dp[x1 - 1][y2][i]))
        elif y1>0 and x1==0:
            k[i] = (dp[x2][y2][i]-dp[x2][y1-1][i])
        else:
            k[i]=dp[x2][y2][i]
        if i>=c:
            if p==0:
                r=i
            else:
                r=(p-1)%(c+1)
        else:
            if p>(c-i):
                r=(p-1-(c-i))%(c+1)
            else:
                r=(p+i)
        ans+=(k[i])*(r)
    print(ans)









