Codeforces Round #427 (Div. 2) Problem C Star sky (Codeforces 835C) - 前缀和

The Cartesian coordinate system is set in the sky. There you can see n stars, the i-th has coordinates (x_i, y_i), a maximum brightness c, equal for all stars, and an initial brightness s_i (0 ≤ s_i ≤ c).

Over time the stars twinkle. At moment 0 the i-th star has brightness s_i. 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 t_i and you will see a rectangle with sides parallel to the coordinate axes, the lower left corner has coordinates (x_1i, y_1i) and the upper right — (x_2i, y_2i). 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 x_i, y_i, s_i (1 ≤ x_i, y_i ≤ 100, 0 ≤ s_i ≤ 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 t_i, x_1i, y_1i, x_2i, y_2i (0 ≤ t_i ≤ 10⁹, 1 ≤ x_1i < x_2i ≤ 100, 1 ≤ y_1i < y_2i ≤ 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.

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　　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 ;
 }