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 java.util.*;
public class Practice{
     public static void main(String args[]){
          Scanner sc = new Scanner(System.in);
          
          int n = sc.nextInt();
          int q = sc.nextInt();
          int c = sc.nextInt();
          
          int xi[] = new int[n];
          int yi[] = new int[n];
          int si[] = new int[n];
          
          int precomputation[][][] = new int[100][100][11];
          
          int starsAtXYwithInitialBrigtness[][][] = new int[100][100][11];
          
          for(int i=0;i<n;i++){
              
               xi[i] = sc.nextInt();
               yi[i] = sc.nextInt();
               si[i] = sc.nextInt();
               
               starsAtXYwithInitialBrigtness[xi[i]-1][yi[i]-1][si[i]] = starsAtXYwithInitialBrigtness[xi[i]-1][yi[i]-1][si[i]] + 1;
               
              
                 
          }
          
          
          for(int i=0;i<100;i++){
              for(int j=0;j<100;j++){
                  for(int k=0;k<11;k++){
                      precomputation[i][j][0] = precomputation[i][j][0] +  (starsAtXYwithInitialBrigtness[i][j][k])*k;
                  }
              }
          }
          
          
          for(int i=0;i<100;i++){
              for(int j=0;j<100;j++){
                  
                  for(int l=0;l<11;l++){
                      
                      if(starsAtXYwithInitialBrigtness[i][j][l]>0){
                        
                          int noOfStarsWithBrightnessL = starsAtXYwithInitialBrigtness[i][j][l];
                          int previousBrightness = l;
                          for(int k=1;k<11;k++){
                              if(previousBrightness+1<=c){
                                  precomputation[i][j][k] = precomputation[i][j][k] + (previousBrightness+1)*noOfStarsWithBrightnessL;
                                  previousBrightness = previousBrightness +1;
                              }else{
                                  precomputation[i][j][k] = precomputation[i][j][k] + 0;
                                  previousBrightness = 0;
                              }
                            
                          }
                      } 
                  }
                  
              }
          }
          
          
          
          
          
          for(int k=0;k<11;k++){
              for(int i=0;i<100;i++){
                  for(int j=1;j<100;j++){
                      precomputation[i][j][k]  = precomputation[i][j][k] + precomputation[i][j-1][k];
                  }
              }
          }
          
           for(int k=0;k<11;k++){
               for(int i=0;i<100;i++){
                    for(int j=0;j<100;j++){
                
                        if(j==0){}
                        else{
                            precomputation[j][i][k] = precomputation[j][i][k] + precomputation[j-1][i][k];
                        }
                    }
                }
            }
          
          
          int ti, x1i, y1i, x2i, y2i,brigtness=0; 
          for(int i=0;i<q;i++){
              brigtness=0;
               ti = sc.nextInt();
               x1i = sc.nextInt();
               y1i = sc.nextInt();
               x2i = sc.nextInt();
               y2i = sc.nextInt();
               
               if(ti>10){
                   ti = ti%(c+1);
               }
               
               int a,b,c1=0;
        if(x1i-1==0){
            a=0;
            c1=0;
        }
        else{
            a = precomputation[(x1i-1)-1][y2i-1][ti];
            if((y1i-1)!=0)
              c1 = precomputation[(x1i-1)-1][(y1i-1)-1][ti];
        }
        if((y1i-1)==0){
            b=0;
            c1=0;
        }else{
            b = precomputation[x2i-1][(y1i-1)-1][ti];
            if((x1i-1)!=0)
                c1 = precomputation[(x1i-1)-1][(y1i-1)-1][ti];
        }
        brigtness = precomputation[x2i-1][y2i-1][ti] - a - b + c1;
               
               System.out.println(brigtness);
          }
     }
}