Problem:
Score : 800 points

Problem Statement

12:17 (UTC): The sample input 1 and 2 were swapped. The error is now fixed. We are very sorry for your inconvenience.

There are N children in AtCoder Kindergarten, conveniently numbered 1 through N. Mr. Evi will distribute C indistinguishable candies to the children.

If child i is given a candies, the child's happiness will become x_i^a, where x_i is the child's excitement level. The activity level of the kindergarten is the product of the happiness of all the N children.

For each possible way to distribute C candies to the children by giving zero or more candies to each child, calculate the activity level of the kindergarten. Then, calculate the sum over all possible way to distribute C candies. This sum can be seen as a function of the children's excitement levels x_1,..,x_N, thus we call it f(x_1,..,x_N).

You are given integers A_i,B_i (1≦i≦N). Find modulo 10^9+7.

Constraints

1≦N≦400

1≦C≦400

1≦A_i≦B_i≦400 (1≦i≦N)

Partial Score

400 points will be awarded for passing the test set satisfying A_i=B_i (1≦i≦N).

Input

The input is given from Standard Input in the following format:

N C
A_1 A_2 ... A_N
B_1 B_2 ... B_N

Output

Print the value of modulo 10^9+7.

Sample Input 1

2 3
1 1
1 1

Sample Output 1

4

This case is included in the test set for the partial score, since A_i=B_i.
We only have to consider the sum of the activity level of the kindergarten where the excitement level of both child 1 and child 2 are 1 (f(1,1)).

If child 1 is given 0 candy, and child 2 is given 3 candies, the activity level of the kindergarten is 1^0*1^3=1.

If child 1 is given 1 candy, and child 2 is given 2 candies, the activity level of the kindergarten is 1^1*1^2=1.

If child 1 is given 2 candies, and child 2 is given 1 candy, the activity level of the kindergarten is 1^2*1^1=1.

If child 1 is given 3 candies, and child 2 is given 0 candy, the activity level of the kindergarten is 1^3*1^0=1.

Thus, f(1,1)=1+1+1+1=4, and the sum over all f is also 4.

Sample Input 2

1 2
1
3

Sample Output 2

14

Since there is only one child, child 1's happiness itself will be the activity level of the kindergarten. Since the only possible way to distribute 2 candies is to give both candies to child 1, the activity level in this case will become the value of f.

When the excitement level of child 1 is 1, f(1)=1^2=1.

When the excitement level of child 1 is 2, f(2)=2^2=4.

When the excitement level of child 1 is 3, f(3)=3^2=9.

Thus, the answer is 1+4+9=14.

Sample Input 3

2 3
1 1
2 2

Sample Output 3

66

Since it can be seen that f(1,1)=4 , f(1,2)=15 , f(2,1)=15 , f(2,2)=32, the answer is 4+15+15+32=66.

Sample Input 4

4 8
3 1 4 1
3 1 4 1

Sample Output 4

421749

This case is included in the test set for the partial score.

Sample Input 5

3 100
7 6 5
9 9 9

Sample Output 5

139123417

Solution in C (GCC 5.4.1) (judge verdict: Accepted)

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

typedef int32_t i32;
typedef int64_t i64;

#define POS(i, j) ((j) * (c + 1) + (i))

void run (void) {
  i32 n, c;
  scanf ("%" SCNi32 "%" SCNi32, &n, &c);
  const i32 m = 400;
  i32 *s = (i32 *) calloc ((c + 1) * (m + 1), sizeof (i32));
  const i32 mod = 1000000007;
  for (i32 i = 1; i <= m; ++i) {
    for (i32 j = 0, t = 1; j <= c; ++j, t = (i64) t * i % mod) {
      s[POS(j, i)] = t;
    }
  }
  for (i32 i = 0; i <= c; ++i) {
    for (i32 j = 1; j <= m; ++j) {
      s[POS(i, j)] = (s[POS(i, j)] + s[POS(i, j - 1)]) % mod;
    }
  }
  i32 *a = (i32 *) calloc (2 * n, sizeof (i32));
  i32 *b = a + n;
  for (i32 i = 0; i < 2 * n; ++i) {
    scanf ("%" SCNi32, a + i);
  }
  i32 *dp = (i32 *) calloc (c + 1, sizeof (i32));
  dp[0] = 1;
  for (i32 i = 0; i < n; ++i) {
    for (i32 j = c; j >= 0; --j) {
      i32 local = 0;
      for (i32 k = 0; k <= j; ++k) {
        local = (local + (i64) (s[POS(k, b[i])] + mod - s[POS(k, a[i] - 1)]) * dp[j - k]) % mod;
      }
      dp[j] = local;
    }
  }
  printf ("%" PRIi32 "\n", dp[c]);
}

int main (void) {
  run();
  return 0;
}