#include <bits/stdc++.h>
using namespace std;
#define int long long
const int MOD = 1e9 + 7;
int addm(int a,int b){ a+=b; if(a>=MOD) a-=MOD; return a; }
int mulm(long long a,long long b){ return (a%MOD)*(b%MOD)%MOD; }
signed main(){
ios::sync_with_stdio(false);
cin.tie(nullptr);
int n;
cin >> n;
vector<int> a(n), b(n);
for(int i=0;i<n;i++) cin >> a[i];
for(int i=0;i<n;i++) cin >> b[i];
// powers of two
vector<int> pw2(n+5,1);
for(int i=1;i<(int)pw2.size();i++) pw2[i] = (pw2[i-1]*2)%MOD;
// distinct levels m we need to sweep
vector<int> levels;
levels.reserve(2*n);
for(int i=0;i<n;i++){ levels.push_back(a[i]); levels.push_back(b[i]); }
sort(levels.begin(), levels.end());
levels.erase(unique(levels.begin(), levels.end()), levels.end());
long long ans = 0;
for(int m : levels){
// ge[i] = # of choices >= m at i; lt[i] = # of choices < m
vector<int> ge(n), lt(n);
for(int i=0;i<n;i++){
ge[i] = (a[i]>=m) + (b[i]>=m);
lt[i] = (a[i]<m) + (b[i]<m);
}
// prefix counts of lt==0 and lt==2 so we can query any interval fast
vector<int> prefZero(n+1,0), prefTwo(n+1,0);
for(int i=0;i<n;i++){
prefZero[i+1] = prefZero[i] + (lt[i]==0);
prefTwo[i+1] = prefTwo[i] + (lt[i]==2);
}
for(int i=1;i<n;i++){
if(ge[i]==0) continue; // right endpoint can't be a wall at level m
for(int j=i-1;j>=0;j--){
if(ge[j]==0) continue; // left endpoint can't be a wall at level m
// middles are indices (j+1 .. i-1)
int zeroCnt = prefZero[i] - prefZero[j+1];
if(zeroCnt>0) continue; // some middle has no <m option -> impossible
int cnt2 = prefTwo[i] - prefTwo[j+1]; // middles with both < m
long long ways = 1;
ways = mulm(ways, ge[j]); // choose >= m at j
ways = mulm(ways, ge[i]); // choose >= m at i
ways = mulm(ways, pw2[cnt2]); // middles with both < m
ways = mulm(ways, pw2[j]); // free left side
ways = mulm(ways, pw2[n-1-i]); // free right side
long long contrib = mulm(i - j - 1, ways);
ans = addm(ans, contrib);
}
}
}
cout << ans % MOD << '\n';
return 0;
}
