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 Java8 (OpenJDK 1.8.0) (judge verdict: Time Limit Exceeded)

//import java.awt.*;
import java.io.*;
import java.math.*;
import java.text.*;
import java.util.*;

public class Main {
  public static void main(String[] args) {
    InputStream inputStream;
    if (args.length > 0 && args[0].equals("devTesting")) {
      try {
        inputStream = new FileInputStream(args[1]);
      } catch(FileNotFoundException e) {
        throw new RuntimeException(e);
      }
    } else {
      inputStream = System.in;
    }
    OutputStream outputStream = System.out;
    InputReader in = new InputReader(inputStream);
    PrintWriter out = new PrintWriter(outputStream);
    TaskE solver = new TaskE();
    solver.solve(1, in, out);
    out.close();
  }
  
  static class TaskE {
    final int ring = (int) 1e9 + 7;
    int n, c;
    int[] ai, bi;
    int[][] state, pow;
    
    void solve(int testNumber, InputReader in, PrintWriter out) {
      n = in.nextInt();
      c = in.nextInt();
      ai = new int[n];
      bi = new int[n];
      for (int i = 0; i < n; ++i) {
        ai[i] = in.nextInt();
      }
      for (int i = 0; i < n; ++i) {
        bi[i] = in.nextInt();
      }
      pow = new int[401][401];
      for (int i = 0; i <= 400; ++i) {
        for (int j = 0; j <= 400; ++j) {
          pow[i][j] = (int) pow(i, j);
        }
      }
      state = new int[n][c + 1];
      for (int[] i : state) {
        Arrays.fill(i, -1);
      }
      long ans = go(n - 1, c);
      out.println(ans);
    }
    
    long go(int pos, int cnd) {
      if (pos < 0) {
        if (cnd == 0) {
          return 1L;
        }
        return 0L;
      }
      if (state[pos][cnd] != -1) {
        return state[pos][cnd];
      }
      long ans = 0L;
      for (int i = 0; i <= cnd; ++i) {
        long x = 0L;
        for (int j = ai[pos]; j <= bi[pos]; ++j) {
          x = add(x, pow[j][i]);
        }
        ans = add(ans, mul(x, go(pos - 1, cnd - i)));
      }
      state[pos][cnd] = (int) ans;
      return ans;
    }
    
    long add(long a, long b) {
      long ans = (a + b) % ring;
      if (ans < 0L) {
        ans += ring;
      }
      return ans;
    }
    
    long mul(long a, long b) {
      return a * b % ring;
    }
    
    long pow(long a, long b) {
      if (a == 0L) {
        return 0L;
      }
      long ans = 1L;
      while (b != 0) {
        if ((b & 1) != 0) {
          ans = mul(ans, a);
        }
        a = mul(a, a);
        b >>>= 1;
      }
      return ans;
    }
    
  }
    
      
  static class InputReader {
    public BufferedReader reader;
    public StringTokenizer tokenizer;

    public InputReader(InputStream stream) {
      reader = new BufferedReader(new InputStreamReader(stream));
      tokenizer = null;
    }

    public String next() {
      while (tokenizer == null || !tokenizer.hasMoreTokens()) {
        try {
          tokenizer = new StringTokenizer(reader.readLine());
        } catch (IOException e) {
          throw new RuntimeException(e);
        }
      }
      return tokenizer.nextToken();
    }
    
    public String nextLine() {
      try {
        return reader.readLine();
      } catch(IOException e) {
        throw new RuntimeException(e);
      }
    }

    public int nextInt() {
      return Integer.parseInt(next());
    }

    public long nextLong() {
      return Long.parseLong(next());
    }
    
    public double nextDouble() {
      return Double.parseDouble(next());
    }

    public boolean hasInput() {
      try {
        if (tokenizer != null && tokenizer.hasMoreTokens()) {
          return true;
        }
        reader.mark(1);
        int ch = reader.read();
        if (ch != -1) {
          reader.reset();
          return true;
        }
        return false;
      } catch(IOException e) {
        throw new RuntimeException(e);
      }
    }
    
  }
}