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\nimport numpy as np\n\nMOD = 10 ** 9 + 7\n\nN,C = map(int,input().split())\nA = [int(x) for x in input().split()]\nB = [int(x) for x in input().split()]\n\n# (i,j) に j^i を入れる\nkth_pow = np.ones((1024, 401), dtype=np.int64)\nrng = np.arange(401, dtype=np.int64)\nfor i in range(1,C+1):\n    kth_pow[i] = kth_pow[i-1] * rng % MOD\nkth_pow_cum = kth_pow.cumsum(axis = 1) % MOD\nkth_pow_cum[C+1:] = 0\n\ndef convolve(A,B,n=32):\n    if n == 8:\n        return np.rint(np.fft.irfft(np.fft.rfft(A) * np.fft.rfft(B))).astype(np.int64)\n    n //= 2\n    M = 1 << n\n    A1 = A // M\n    A2 = A - M * A1\n    B1 = A // M\n    B2 = B - M * B1\n    X = convolve(A1,B1,n)\n    Y = convolve(A1-A2,B1-B2,n)\n    Z = convolve(A2,B2,n)\n    return (X * (M * M % MOD) + (X + Z - Y) * M + Z) % MOD\n\ndp = np.zeros(1024, dtype=np.int64) # これまで配った個数、合計点\ndp[0] = 1\nfor a,b in zip(A,B):\n    arr = kth_pow_cum[:,b] - kth_pow_cum[:,a-1]\n    dp = convolve(dp,arr)\n    dp[C+1:] = 0\n\nanswer = dp[C]\nprint(answer)'
_co = compile(_CODE, USER_FILENAME, "exec")
exec(_co, ns, ns)

```

Measured memory_bytes: 15949667