#include <bits/stdc++.h>
using namespace std;
#define dbg(x) //x
#define prt(x) dbg(cerr << x)
#define pv(x) dbg(cerr << #x << " = " << x << '\n')
#define parr(x) dbg(cerr << #x << " = "; for (auto y : x) cerr << y << ' '; cerr << '\n';)
#define parr2d(x) dbg(cerr << #x << " = \n"; for (auto _ : x) {parr(_);} cerr << '\n';)
/*
1
bitmask
2
for each pair of walls
calculate the sum of all possible sums of elements between these
such that all of them are less than a certain value
for this, doing all ranges & all values will work
3
4
5
let i be the end of a segment
dp[i] = the answer
ways[i] = # of ways so that h[i] = 2
for i, for all j <= i - 2
dp[i] += dp[j] + ways[j] * (i - j - 1)
ways[i] += ways[j]
*/
const long long mod = 1e9 + 7;
int main () {
ios::sync_with_stdio(0); cin.tie(0);
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];
}
vector<long long> dp(n, 0), ways(n, 0);
ways[0] = 1;
ways[1] = 2;
long long dp_pref = 0, ways_pref = 1, ways_sum = 1;
for (int i = 2; i < n; i++) {
pv(i); pv(ways_sum);
dp[i] = (dp_pref + ways_sum) % mod;
dp[i] += dp[i - 1]; dp[i] %= mod;
ways[i] = (ways[i - 1] * 2) % mod;
dp_pref += dp[i - 1]; dp_pref %= mod;
ways_pref += ways[i - 1]; ways_pref %= mod;
ways_sum += ways_pref; ways_sum %= mod;
}
parr(ways); parr(dp);
dp_pref += dp[n - 1]; dp_pref %= mod;
cout << dp_pref << '\n';
}
