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

a=[int(i) for i in sys.stdin.readline().split()]
cc=a[2]
st=[]
s=[]
ch=[[[0,0,0,0,0,0,0,0,0,0,0] for i in range(101)] for j in range(101)]
for i in range(a[0]):
    c=[int(j) for j in sys.stdin.readline().split()]
    st.append((c[0],c[1]))
    s.append(c[2])
    ch[c[0]][c[1]][c[2]]+=1
dp=[[[0,0,0,0,0,0,0,0,0,0,0] for i in range(102)] for j in range(102)]
for i in range(1,102):
    for j in range(1,102):
        for k in range(11):
            dp[i][j][k] = dp[i-1][j][k] + dp[i][j-1][k] - dp[i-1][j-1][k]
        for k in range(11):
            dp[i][j][k] +=ch[i-1][j-1][k]
anss=[]
for i in range(a[1]):
    ans=0
    l=[int(j) for j in sys.stdin.readline().split()]
    pp=l[0]
    i1=l[1]
    j1=l[2]
    i2=l[3]
    j2=l[4]
    count = [dp[i2+1][j2+1][k] + dp[i1][j1][k] - dp[i2+1][j1][k] - dp[i1][j2+1][k] for k in range(11)]
    t=l[0]
    for m in range(11):
        lk=(m+t)%(cc+1)
        ans+=(count[m]*lk)
        #print(m,lk,ans)
    anss.append(str(ans))
print("\n".join(anss))
    #break