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 <bits/stdc++.h>
using namespace std;

typedef long long ll;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;

const int MAXH = 18;
const int MAXN = (1<<18);
const ll MOD = 1e9+7;

int H, P[MAXN+10];
ll A[MAXN+10], ans;

void solve(int node, int tl, int tr)
{
	if(tl==tr) return;
	int mid=tl+tr>>1;
	solve(node*2, tl, mid);
	solve(node*2+1, mid+1, tr);

	vector<int> LV1, LV2, RV1, RV2;
	for(int i=tl; i<=mid; i++) LV1.push_back((1<<(H-1))-1+i);
	for(int i=tl; i<=mid; i++) LV2.push_back(P[i]+(1<<(H-1))-1);
	for(int i=mid+1; i<=tr; i++) RV1.push_back((1<<(H-1))-1+i);
	for(int i=mid+1; i<=tr; i++) RV2.push_back(P[i]+(1<<(H-1))-1);

	for(int i=0; i<RV1.size(); i++)
	{
		int u=RV1[i]; ll val=node;
		while(u!=node)
		{
			val=val*u%MOD;
			u/=2;
		}
		int v=RV2[i];
		while(v)
		{
			val=val*v%MOD;
			A[v]=(A[v]+val)%MOD;
			v/=2;
		}
	}
	for(int i=0; i<LV1.size(); i++)
	{
		int u=LV1[i]; ll val=1;
		while(u!=node)
		{
			val=val*u%MOD;
			u/=2;
		}
		int v=LV2[i];
		ll bef=0;
		while(v)
		{
			ans+=val*(A[v]-bef*v%MOD);
			ans=(ans%MOD+MOD)%MOD;
			bef=A[v];
			val=val*v%MOD;
			v/=2;
		}
	}
	for(int i=0; i<RV1.size(); i++)
	{
		int v=RV2[i];
		while(v)
		{
			A[v]=0;
			v/=2;
		}
	}
}

int main()
{
	scanf("%d", &H);
	for(int i=1; i<=(1<<(H-1)); i++) scanf("%d", &P[i]);

	solve(1, 1, (1<<(H-1)));
	printf("%lld\n", ans);
}