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.


  题目大意 天空中有一些星星,每个星星有一个初始亮度,如果一个星星的初始亮度为s, 那么在时刻t, 它的亮度为(s + t) % (c + 1)。有q个询问,询问在某一时刻天空中某个矩形内所有星星的亮度和。

  x, y很小,c很小,而且有趣的是10 * 100 * 100 = 100000,标准cf数据范围。

  所以考虑对每种星星的初始亮度搞一个前缀和。这样对于某一时刻,你可以算出某个矩形内亮度为x的星星数目。

  于是这道题就很简单了。。

  值得高兴的是,终于在考试的时候把C题A掉了。。。好开心。。一直认为C题有毒,每次都会做,每次都挂。

Code

 /**
* Codeforces
* Problem#835C
* Accepted
* Time:156ms
* Memory:3700k
*/
#include <bits/stdc++.h>
#ifndef WIN32
#define Auto "%lld"
#else
#define Auto "%I64d"
#endif
using namespace std;
typedef bool boolean;
const signed int inf = (signed)((1u << ) - );
const signed long long llf = (signed long long)((1ull << ) - );
const double eps = 1e-;
const int binary_limit = ;
#define smin(a, b) a = min(a, b)
#define smax(a, b) a = max(a, b)
#define max3(a, b, c) max(a, max(b, c))
#define min3(a, b, c) min(a, min(b, c))
template<typename T>
inline boolean readInteger(T& u){
char x;
int aFlag = ;
while(!isdigit((x = getchar())) && x != '-' && x != -);
if(x == -) {
ungetc(x, stdin);
return false;
}
if(x == '-'){
x = getchar();
aFlag = -;
}
for(u = x - ''; isdigit((x = getchar())); u = (u << ) + (u << ) + x - '');
ungetc(x, stdin);
u *= aFlag;
return true;
} int n, q, c;
int xs[], ys[], ss[];
int sum[][][]; inline void init() {
scanf("%d%d%d", &n, &q, &c);
for(int i = ; i <= n; i++) {
scanf("%d%d%d", xs + i, ys + i, ss + i);
}
} inline void getPreSum() {
memset(sum, , sizeof(sum));
for(int i = ; i <= n; i++) {
sum[xs[i]][ys[i]][ss[i]]++;
}
for(int i = ; i <= ; i++) {
for(int j = ; j <= ; j++) {
for(int k = ; k <= ; k++)
sum[i][j][k] += sum[i - ][j][k] + sum[i][j - ][k] - sum[i - ][j - ][k];
}
}
} inline int getAns(int x, int y, int t0) {
int rt = ;
for(int i = ; i <= ; i++)
rt += (sum[x][y][i]) * ((i + t0) % (c + ));
return rt;
} inline void solve() {
int t0, x0, y0, x1, y1;
while(q--) {
scanf("%d%d%d%d%d", &t0, &x0, &y0, &x1, &y1);
printf("%d\n", getAns(x1, y1, t0) - getAns(x0 - , y1, t0) - getAns(x1, y0 - , t0) + getAns(x0 - , y0 - , t0));
}
} int main() {
init();
getPreSum();
solve();
return ;
}

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