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-----
2 3
1 1
1 1

-----Sample Output-----
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.

```python

USER_FILENAME = "<user_code>"
import sys, json, builtins
import math, heapq, bisect, itertools, functools, collections, statistics, fractions, decimal, operator, array, random, re, string, os, pathlib
try:
    import resource as _res
except Exception:
    _res = None
from collections import Counter, defaultdict, deque, OrderedDict
from functools import lru_cache, reduce
from itertools import accumulate, permutations, combinations, product
from math import gcd, ceil, floor, sqrt
from decimal import Decimal
try:
    from typing import *
except Exception:
    pass
try:
    sys.setrecursionlimit(1_000_000)
except Exception:
    pass
try:
    import site as _site
    try:
        _sys_sites = _site.getsitepackages() if hasattr(_site, "getsitepackages") else []
        for _p in _sys_sites:
            if _p and _p not in sys.path:
                sys.path.append(_p)
    except Exception:
        pass
    try:
        _usp = _site.getusersitepackages()
        if isinstance(_usp, str):
            _usp = [_usp]
        for _p in _usp:
            if _p and _p not in sys.path:
                sys.path.append(_p)
    except Exception:
        pass
except Exception:
    pass

ns = {"__name__": "__main__"}
for _k, _v in list(globals().items()):
    if not _k.startswith("_"):
        ns[_k] = _v
_CODE = 'import sys\ninput = sys.stdin.readline\n\n# modを取りながらべき乗する\ndef power_func(a,n,mod):\n    bi=str(format(n,"b"))#2進表現に\n    res=1\n    for i in range(len(bi)):\n        res=(res*res) %mod\n        if bi[i]=="1":\n            res=(res*a) %mod\n    return res\n\ndef main():\n    N, C = map(int, input().split())\n    A = list(map(int, input().split()))\n    B = list(map(int, input().split()))\n    mod = 10**9+7\n    L = 400\n\n    dp = [[0]*(L+1) for _ in range(C+1)]\n    count = B[0] - A[0] + 1\n    for c in range(C+1):\n        a, b = A[0], B[0]\n        dp[c][0] = power_func(a, c, mod)\n        for k in range(1, b-a+1):\n            dp[c][k] = (dp[c][k-1] + power_func(a+k, c, mod)) % mod\n        dp[c][L] = dp[c][b-a]\n\n    for i, (a, b) in enumerate(zip(A, B)):\n        if i == 0: continue\n        for k in range(b-a+1):\n            dp[0][k] = count * (k+1) % mod\n        dp[0][L] = dp[0][b-a]\n        for c in range(1, C+1):\n            dp[c][0] = (dp[c][L] + a*dp[c-1][0]) % mod\n            for k in range(1, b-a+1):\n                R = dp[c][k-1] + dp[c][L] + (a+k)*(dp[c-1][k]-dp[c-1][k-1])\n                if R < 0: R += mod\n                dp[c][k] = R%mod\n            dp[c][L] = dp[c][b-a]\n        count = (count * (b-a+1)) % mod\n\n    print(dp[C][L])\n\n\ndef __starting_point():\n    main()\n__starting_point()'
_co = compile(_CODE, USER_FILENAME, "exec")
exec(_co, ns, ns)

```

Measured memory_bytes: 21221