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 <cstdio>
#include <cmath>
#include <ctime>
#include <cstdlib>
#include <cassert>
#include <vector>
#include <list>
#include <stack>
#include <queue>
#include <deque>
#include <map>
#include <set>
#include <bitset>
#include <string>
#include <algorithm>
#include <utility>
#define llint long long
#define inf 1e18
#define rep(x, s, t) for(llint (x) = (s); (x) < (t); (x)++)
#define Rep(x, s, t) for(llint (x) = (s); (x) <= (t); (x)++)
#define chmin(x, y) (x) = min((x), (y))
#define chmax(x, y) (x) = max((x), (y))
#define mod 1000000007

using namespace std;
typedef pair<llint, llint> P;

llint n, N, M;
llint p[1<<17];
llint a[1<<18], ainv[1<<18], b[1<<18];
llint c[1<<18], d[1<<18];
vector<llint> vec;
llint ans = 0;

llint modpow(llint a, llint n)
{
	if(n == 0) return 1;
	if(n % 2){
		return ((a%mod) * (modpow(a, n-1)%mod)) % mod;
	}
	else{
		return modpow((a*a)%mod, n/2) % mod;
	}
}

void calc2(llint l, llint r, llint c[])
{
	for(int i = l; i <= r; i++){
		llint v = p[i-M]+M;
		while(v){
			c[v] += b[i], c[v] %= mod;
			vec.push_back(v);
			v /= 2;
		}
	}
}

void calc(llint x, llint l, llint r)
{
	llint m = (l+r)/2;
	
	vec.clear();
	calc2(l, m, c);
	calc2(m+1, r, d);
	
	sort(vec.begin(), vec.end());
	vec.erase(unique(vec.begin(), vec.end()), vec.end());
	reverse(vec.begin(), vec.end());
	
	llint ret = 0;
	for(int i = 0; i < vec.size(); i++){
		llint v = vec[i];
		if(v*2 < N){
			llint tmp = c[v*2] * d[v*2+1] % mod;
			tmp += d[v*2] * c[v*2+1] % mod, tmp %= mod;
			tmp *= ainv[v], tmp %= mod;
			ret += tmp, ret %= mod;
		}
	}
	for(int i = 0; i < vec.size(); i++){
		llint v = vec[i];
		c[v] = d[v] = 0;
	}
	
	ret *= ainv[x], ret %= mod;
	ans += ret, ans %= mod;
	
	if(x*2 < M) calc(2*x, l, m), calc(2*x+1, m+1, r);
}

int main(void)
{
	ios::sync_with_stdio(0);
	cin.tie(0);
	
	cin >> n;
	N = 1<<n, M = 1<<(n-1);
	for(int i = 0; i < M; i++) cin >> p[i], p[i]--;
	
	a[1] = 1;
	for(int i = 2; i < N; i++) a[i] = a[i/2] * i % mod;
	for(int i = 1; i < N; i++) ainv[i] = modpow(a[i], mod-2) * modpow(a[i], mod-2) % mod * i % mod;
	for(int i = 0; i < M; i++) b[M+i] = a[M+i] * a[M+p[i]] % mod;
	
	calc(1, M, N-1);
	cout << ans << endl;
	
	return 0;
}