p01769 Hopping Hearts
Problem statement

N-winged rabbit is on a balance beam of length L-1. The initial position of the i-th rabbit is the integer x_i, which satisfies 0 ≤ x_ {i} \ lt x_ {i + 1} ≤ L−1. The coordinates increase as you move to the right. Any i-th rabbit can jump to the right (ie, move from x_i to x_i + a_i) any number of times, just a distance a_i. However, you cannot jump over another rabbit or enter a position below -1 or above L. Also, at most one rabbit can jump at the same time, and at most one rabbit can exist at a certain coordinate.

How many possible states of x_ {0},…, x_ {N−1} after starting from the initial state and repeating the jump any number of times? Find by the remainder divided by 1 \, 000 \, 000 \, 007.

input

The input is given in the following format.


N L
x_ {0}… x_ {N−1}
a_ {0}… a_ {N−1}


Constraint

* All inputs are integers
* 1 \ ≤ N \ ≤ 5 \,000
* N \ ≤ L \ ≤ 5 \,000
* 0 \ ≤ x_ {i} \ lt x_ {i + 1} \ ≤ L−1
* 0 \ ≤ a_ {i} \ ≤ L−1



output

Print the answer in one line.

sample

Sample input 1


13
0
1


Sample output 1


3


If 1/0 is used to express the presence / absence of a rabbit, there are three ways: 100, 010, and 001.

Sample input 2


twenty four
0 1
1 2


Sample output 2


Four


There are four ways: 1100, 1001, 0101, 0011.

Sample input 3


10 50
0 1 2 3 4 5 6 7 8 9
1 1 1 1 1 1 1 1 1 1


Sample output 3


272278100


The binomial coefficient C (50,10) = 10 \, 272 \, 278 \, 170, and the remainder obtained by dividing it by 1 \, 000 \, 000 \, 007 is 272 \, 278 \, 100.





Example

Input

1 3
0
1


Output

3
```python
def main():
  MOD = 1000000007
  n, l = map(int, input().split())
  xlst = map(int, input().split())
  alst = map(int, input().split())
  can_use = []
  for x, a in zip(xlst, alst):
    if a == 0:s = {x}
    else:s = {k for k in range(x, l, a)}
    can_use.append(s)
  
  dp = [[0] * l for _ in range(n)]
  for j in range(l):
    dp[0][j] = dp[0][j - 1] + int(j in can_use[0])
  
  for i in range(1, n):
    acc = 0
    dpi = dp[i]
    dpi1 = dp[i - 1]
    st = can_use[i]
    for j in range(1, l):
      if j in st:acc = (acc + dpi1[j - 1]) % MOD
      dpi[j] = acc
  print(dp[n - 1][l - 1])

main() 

```