AtCoder Regular Contest 059
E
Children and Candies
AtCoder
⌀ 2189
原题
← DF →
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Problem Statement

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

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

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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.

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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 $f(x_1,..,x_N)$.

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

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Constraints

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  • $1≦N≦400$
  • $1≦C≦400$
  • $1≦A_i≦B_i≦400 (1≦i≦N)$
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Partial Score

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

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

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

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Output

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Print the value of image modulo $10^9+7$.

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样例

Input #1
2 3
1 1
1 1
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 $4$.
Input #2
1 2
1
3
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 $1+4+9=14$.
Input #3
2 3
1 1
2 2
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 $4+15+15+32=66$.
Input #4
4 8
3 1 4 1
3 1 4 1
Output #4
421749
This case is included in the test set for the partial score.
Input #5
3 100
7 6 5
9 9 9
Output #5
139123417
题目信息
题号
arc059_c
时间限制
4 s
内存限制
256 MB

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