Inspired by the tv series Stranger Things, bear Limak is going for a walk between two mirror worlds.

There are two perfect binary trees of height H, each with the standard numeration of vertices from 1 to 2^H-1. The root is 1 and the children of x are 2 \cdot x and 2 \cdot x + 1.

Let L denote the number of leaves in a single tree, L = 2^{H-1}.

You are given a permutation P_1, P_2, \ldots, P_L of numbers 1 through L. It describes L special edges that connect leaves of the two trees. There is a special edge between vertex L+i-1 in the first tree and vertex L+P_i-1 in the second tree.

graph for sample1

drawing for the first sample test, permutation P = (2, 3, 1, 4), special edges in green

Let's define product of a cycle as the product of numbers in its vertices. Compute the sum of products of all simple cycles that have exactly two special edges, modulo (10^9+7).

A simple cycle is a cycle of length at least 3, without repeated vertices or edges.

Constraints

* 2 \leq H \leq 18
* 1 \leq P_i \leq L where L = 2^{H-1}
* P_i \neq P_j (so this is a permutation)

Input

Input is given from Standard Input in the following format (where L = 2^{H-1}).


H
P_1 P_2 \cdots P_L


Output

Compute the sum of products of simple cycles that have exactly two special edges. Print the answer modulo (10^9+7).

Examples

Input

3
2 3 1 4


Output

121788


Input

2
1 2


Output

36


Input

5
6 14 15 7 12 16 5 4 11 9 3 10 8 2 13 1


Output

10199246
#include<iostream>
#include<string>
#include<cstdio>
#include<vector>
#include<cmath>
#include<algorithm>
#include<functional>
#include<iomanip>
#include<queue>
#include<ciso646>
#include<random>
#include<map>
#include<set>
#include<bitset>
#include<stack>
#include<unordered_map>
#include<utility>
#include<cassert>
#include<complex>
#include<numeric>
using namespace std;

//#define int long long
typedef long long ll;

typedef unsigned long long ul;
typedef unsigned int ui;
constexpr ll mod = 1000000007;
const ll INF = mod * mod;
typedef pair<int, int>P;
#define stop char nyaa;cin>>nyaa;
#define rep(i,n) for(int i=0;i<n;i++)
#define per(i,n) for(int i=n-1;i>=0;i--)
#define Rep(i,sta,n) for(int i=sta;i<n;i++)
#define rep1(i,n) for(int i=1;i<=n;i++)
#define per1(i,n) for(int i=n;i>=1;i--)
#define Rep1(i,sta,n) for(int i=sta;i<=n;i++)
#define all(v) (v).begin(),(v).end()
typedef pair<ll, ll> LP;
typedef double ld;
typedef pair<ld, ld> LDP;
const ld eps = 1e-12;
const ld pi = acos(-1.0);

ll mod_pow(ll x, ll n, ll m) {
	ll res = 1;
	while (n) {
		if (n & 1)res = res * x % m;
		x = x * x % m; n >>= 1;
	}
	return res;
}
struct modint {
	ll n;
	modint() :n(0) { ; }
	modint(ll m) :n(m) {
		if (n >= mod)n %= mod;
		else if (n < 0)n = (n % mod + mod) % mod;
	}
	operator int() { return n; }
};
bool operator==(modint a, modint b) { return a.n == b.n; }
modint operator+=(modint& a, modint b) { a.n += b.n; if (a.n >= mod)a.n -= mod; return a; }
modint operator-=(modint& a, modint b) { a.n -= b.n; if (a.n < 0)a.n += mod; return a; }
modint operator*=(modint& a, modint b) { a.n = ((ll)a.n * b.n) % mod; return a; }
modint operator+(modint a, modint b) { return a += b; }
modint operator-(modint a, modint b) { return a -= b; }
modint operator*(modint a, modint b) { return a *= b; }
modint operator^(modint a, ll n) {
	if (n == 0)return modint(1);
	modint res = (a * a) ^ (n / 2);
	if (n % 2)res = res * a;
	return res;
}

ll inv(ll a, ll p) {
	return (a == 1 ? 1 : (1 - p * inv(p % a, a)) / a + p);
}
modint operator/(modint a, modint b) { return a * modint(inv(b, mod)); }

const int max_n = 1 << 17;
modint fact[max_n], factinv[max_n];
void init_f() {
	fact[0] = modint(1);
	for (int i = 0; i < max_n - 1; i++) {
		fact[i + 1] = fact[i] * modint(i + 1);
	}
	factinv[max_n - 1] = modint(1) / fact[max_n - 1];
	for (int i = max_n - 2; i >= 0; i--) {
		factinv[i] = factinv[i + 1] * modint(i + 1);
	}
}
modint comb(int a, int b) {
	if (a < 0 || b < 0 || a < b)return 0;
	return fact[a] * factinv[b] * factinv[a - b];
}

ll gcd(ll a, ll b) {
	if (a < b)swap(a, b);
	while (b) {
		ll r = a % b; a = b; b = r;
	}
	return a;
}

modint val[1 << 20];
void solve() {
	int n; cin >> n;
	int l = (1 << (n - 1));
	vector<int> p(l);
	rep(i, l) {
		cin >> p[i]; p[i]--;
	}
	modint ans = 0;
	per(i, l - 1) {
		modint memo = ans;
		int lh = 2 * i + 1;
		int rh = 2 * i + 2;
		int copl = lh, copr = rh;
		function<void(int, modint)> dfs1 = [&](int k, modint s) {
			if (k >= l - 1) {
				int id = k - (l - 1);
				int loc = p[id];
				loc += l - 1;
				s *= (loc + 1);
				val[loc] = s;
				//cout << " aaa " << loc << "\n";
				while (loc > 0) {
					loc = (loc - 1) / 2;
					s *= (loc + 1);
					val[loc] += s;
					//cout << " aaa " << loc << "\n";
				}
			}
			else {
				dfs1(2 * k + 1, s * (modint)(2 * k + 2));
				dfs1(2 * k + 2, s * (modint)(2 * k + 3));
			}
		};
		dfs1(copl, (copl + 1));

		function<void(int, modint)> dfs2 = [&](int k, modint s) {
			if (k >= l - 1) {
				int id = k - (l - 1);
				int loc = p[id];
				loc += l - 1;
				//cout << "??? " << loc << "\n";
				s *= (loc + 1);
				modint pre = 0;
				while (loc > 0) {
					loc = (loc - 1) / 2;
					pre *= (loc + 1);
					//cout << "hoge " << loc<<" "<<val[loc] << " " << s << "\n";
					ans += (val[loc]-pre) * s;
					pre = val[loc];
					s *= (loc + 1);
				}
			}
			else {
				dfs2(2 * k + 1, s * (modint)(2 * k + 2));
				dfs2(2 * k + 2, s * (modint)(2 * k + 3));
			}
		};
		dfs2(copr, (modint)(i+1)*(modint)(copr + 1));

		function<void(int)> dfs3 = [&](int k) {
			if (k >= l - 1) {
				int id = k - (l - 1);
				int loc = p[id];
				loc += l - 1;
				val[loc] = 0;
				//cout << " bbb " << loc << "\n";
				while (loc > 0) {
					loc = (loc - 1) / 2;
					val[loc] = 0;
					//cout << " bbb " << loc << "\n";
				}
			}
			else {
				dfs3(2 * k + 1);
				dfs3(2 * k + 2);
			}
		};
		dfs3(copl);
		//cout << i << " " << ans - memo << "\n";
		/*rep(j, 2 * l - 1) {
			cout << i << " " << j << " " << val[j] << "\n";
		}*/
	}
	cout << ans << "\n";
}




signed main() {
	ios::sync_with_stdio(false);
	cin.tie(0);
	//cout << fixed << setprecision(10);
	//init_f();
	//expr();
	//int t; cin >> t; rep(i, t)
	solve();
	return 0;
}
