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 C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include<bits/stdc++.h>
//ios::sync_with_stdio(false);cin.tie(0);
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int,int> pii;
typedef pair<pii,int> ppii;
typedef pair<int,pii> pipi;
typedef pair<ll,ll> pll;
typedef pair<pll,ll> ppll;
typedef pair<ll,pll> plpl;
typedef tuple<ll,ll,ll> tl;
ll mod=1000000007;
ll mod2=998244353;
ll mod3=1000003;
ll mod4=998244853;
ll inf=1000000000000000000;
double pi=2*acos(0);
#define rep(i,m,n) for(ll i=m;i<n;i++)
#define rrep(i,n,m) for(ll i=n;i>=m;i--)
int dh[4]={1,-1,0,0};
int dw[4]={0,0,1,-1};
int ddh[8]={-1,-1,-1,0,0,1,1,1};
int ddw[8]={-1,0,1,-1,1,-1,0,1};
ll lmax(ll a,ll b){
    if(a<b)return b;
    else return a;
}
ll lmin(ll a,ll b){
    if(a<b)return a;
    else return b;
}
ll gcd(ll a,ll b){
    if(a<b)swap(a,b);
    if(b==0)return a;
    if(a%b==0)return b;
    return gcd(b,a%b);
}
ll Pow(ll n,ll k){
    ll ret=1;
    ll now=n;
    while(k>0){
        if(k&1)ret*=now;
        now*=now;
        k/=2;
    }
    return ret;
}
ll gya[1000010];
ll kai[1000010];
ll beki(ll n,ll k,ll md){
  ll ret=1;
  ll now=n;
  while(k>0){
    if(k%2==1){
      ret*=now;
      ret%=md;
    }
    now*=now;
    now%=md;
    k/=2;
  }
  return ret;
}
ll gyaku(ll n,ll md){
  return beki(n,md-2,md);
}
void nckinit(ll n,ll md){
  kai[0]=1;
  kai[1]=1;
  for(int i=2;i<=n;i++){
    kai[i]=kai[i-1]*i;
    kai[i]%=md;
  }
  gya[n]=gyaku(kai[n],md);
  for(int i=n-1;i>=1;i--){
    gya[i]=gya[i+1]*(i+1);
    gya[i]%=md;
  }
  gya[0]=1;
}
ll nck(ll n,ll k,ll md){
  if(k<0)return 0;
  if(k==0||n==k)return 1;
  ll ret=kai[n];
  ret*=gya[n-k];
  ret%=md;
  ret*=gya[k];
  ret%=md;
  return ret;
}
ll npk(ll n,ll k,ll md){
  if(k<0)return 0;
  if(k==0)return 1;
  ll ret=kai[n];
  ret*=gya[n-k];
  ret%=md;
  return ret;
}
int main(){
    ios::sync_with_stdio(false);cin.tie(0);
    ll n,c;cin>>n>>c;
    ll sum[401][401];
    rep(i,0,401){
        sum[i][0]=0;
        rep(j,1,401){
            sum[i][j]=(sum[i][j-1]+beki(j,i,mod))%mod;
        }
    }
    /*rep(i,0,c+1){
        rep(j,0,c+1)cout<<sum[i][j]<<" ";
        cout<<endl;
    }*/
    ll dp[n+1][c+1];
    memset(dp,0,sizeof dp);
    dp[0][0]=1;
    ll cc[n],d[n];
    rep(i,0,n)cin>>cc[i];
    rep(i,0,n)cin>>d[i];
    rep(i,1,n+1){
        ll a=cc[i-1];
        ll b=d[i-1];
        rep(j,0,c+1){
            rep(k,0,j+1){
                (dp[i][j]+=dp[i-1][k]*((sum[j-k][b]-sum[j-k][a-1]
                                        +mod)%mod))%=mod;
            }
        }
    }
    cout<<dp[n][c]<<endl;
   /* rep(i,0,n+1){
        rep(j,0,c+1)cout<<dp[i][j]<<" ";
        cout<<endl;
    }*/
}