Problem:
Score : 300 points

Problem Statement

The season for Snuke Festival has come again this year. First of all, Ringo will perform a ritual to summon Snuke. For the ritual, he needs an altar, which consists of three parts, one in each of the three categories: upper, middle and lower.

He has N parts for each of the three categories. The size of the i-th upper part is A_i, the size of the i-th middle part is B_i, and the size of the i-th lower part is C_i.

To build an altar, the size of the middle part must be strictly greater than that of the upper part, and the size of the lower part must be strictly greater than that of the middle part. On the other hand, any three parts that satisfy these conditions can be combined to form an altar.

How many different altars can Ringo build? Here, two altars are considered different when at least one of the three parts used is different.

Constraints

1 \leq N \leq 10^5

1 \leq A_i \leq 10^9(1\leq i\leq N)

1 \leq B_i \leq 10^9(1\leq i\leq N)

1 \leq C_i \leq 10^9(1\leq i\leq N)

All input values are integers.

Input

Input is given from Standard Input in the following format:

N
A_1 ... A_N
B_1 ... B_N
C_1 ... C_N

Output

Print the number of different altars that Ringo can build.

Sample Input 1

2
1 5
2 4
3 6

Sample Output 1

3

The following three altars can be built:

Upper: 1-st part, Middle: 1-st part, Lower: 1-st part

Upper: 1-st part, Middle: 1-st part, Lower: 2-nd part

Upper: 1-st part, Middle: 2-nd part, Lower: 2-nd part

Sample Input 2

3
1 1 1
2 2 2
3 3 3

Sample Output 2

27

Sample Input 3

6
3 14 159 2 6 53
58 9 79 323 84 6
2643 383 2 79 50 288

Sample Output 3

87

Solution in C (GCC 5.4.1) (judge verdict: Wrong Answer)

#include<stdio.h>
#include<stdlib.h>

int *binary_search(int *array, size_t size, int value){
 int lower=0;
 int upper=(int)size - 1;

 while(lower <= upper){
  int mid = (lower + upper) /2;

  if( array[mid] == value){
    return &array[mid];
  }else if(array[mid] < value){
    lower = mid + 1;
  }else if(array[mid] > value){
    upper = mid - 1;
  }
 }

  return NULL;
}

int compare(int a, int b){
  if(a < b){
    return -1;
  }else if(a > b){
    return 1;
  }
  return 0;
}

//イテレーターを返す
int lower_bound(int *array, int len, int value){
  int l=0, r=len - 1;

  while(l <= r){
    int mid = l + (r-l)/2;
    int res = compare(array[mid], value);
    if(res == -1){
      l = mid + 1;
    }else{
      r = mid - 1;
    }
  }

  return l;
}

int upper_bound(int *array, int len, int value){
  int l=0, r=len-1;

  while(l <= r){
    int mid = l + (r-l)/2;
    int res= compare(array[mid], value);
    if(res <= 0) l= mid +1;
      else r = mid -1;
  }

  return l;
}

void merge_sort (int array[], int left, int right) {
  int i, j, k, mid;
  int work[100000];  //work array
  if (left < right) {
    mid = (left + right)/2;
    merge_sort(array, left, mid);  //sort left
    merge_sort(array, mid+1, right);  //sort right
    for (i = mid; i >= left; i--) { work[i] = array[i]; } //left section
    for (j = mid+1; j <= right; j++) {
      work[right-(j-(mid+1))] = array[j]; //right section
    }
    i = left; j = right;
    for (k = left; k <= right; k++) {
      if (work[i] < work[j]) { array[k] = work[i++]; }
      else                   { array[k] = work[j--]; }
    }
  }
}


int main(void){
  int n;
  int *a,*b,*c;

  scanf("%d", &n);
  a = (int *)malloc(sizeof(int) * n); 
  b = (int *)malloc(sizeof(int) * n); 
  c = (int *)malloc(sizeof(int) * n); 

  int temp;
  for(size_t i = 0; i < n;i++){
    scanf("%d", &temp);
    *(a+i) = temp;
  }
  merge_sort(a, 0, n-1);
  for(size_t i = 0; i < n;i++){
    scanf("%d", &temp);
    *(b+i) = temp;
  }
  merge_sort(b, 0, n-1);
  for(size_t i = 0; i < n;i++){
    scanf("%d", &temp);
    *(c+i) = temp;
  }
  merge_sort(c, 0, n-1);
  for(size_t i =0;i<n;i++){
    //printf("%d ", a[i]);
  }
  //printf("\n");
  for(size_t i =0;i<n;i++){
    //printf("%d ", b[i]);
  }
  //printf("\n");
  for(size_t i =0;i<n;i++){
    //printf("%d ", c[i]);
  }
  //printf("\n");

  //上部の数字を選択、上部より大きい中部を選択、中部より大きい下部を選択
  int up,mid,low;
  int count=0;
  int add=0;
  for(size_t i = 0;i < n;i++){
    //中央を基準とする
    mid = b[i];
    low = upper_bound(c, n, mid);
    up = lower_bound(a, n, mid);
    //printf("low:%d\n", n - low);
    //printf("up:%d\n", up);
    count += (n -low) * up;
  }
  printf("%d\n",count);
  return 0;
}