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;
#define rep(i,a,b) for(int i=(a);i<=(b);++i)
typedef long long LL;
const LL P = 1e9+7;
const int N = (1<<18)+5;

int h, n, x, a[N], ans = 0;
LL prod[N], sumprod[N];

LL pw(LL x, int n){
	LL res = 1;
	for(;n;n>>=1,x=x*x%P) if(n&1) res=res*x%P;
	return res;
}

void workl(int l, int r, LL div, bool flag){
	rep(i,l,r){
		LL x = prod[i] * div % P * a[i] % P;
		for(LL y = a[i]>>1; y; y>>=1){
			x = x * y % P;
			if (flag) (sumprod[y] += x) %= P;
			else sumprod[y] = 0;
		}
	}
}

LL workr(int l, int r){
	LL res = 0;
	rep(i,l,r){
		LL x = prod[i] * a[i] % P, done = 0;
		for(LL y = a[i]>>1; y; y>>=1){
			res = (res + (sumprod[y] - done) * x) % P;
			done = sumprod[y] * (y>>1) % P;
			x = x * y % P;
		}
	}
	return (res+P) % P;
}

int main(){
	scanf("%d",&h);
	n = 1<<h-1;
	rep(i,n,2*n-1) scanf("%d",&x), a[i] = n+x-1;
	
	prod[1] = 1;
	rep(i,2,n*2-1) prod[i] = prod[i>>1] * i % P;
	
	rep(k,1,h-1) rep(i,1<<k-1,(1<<k)-1){
		int l = i<<h-k, r = (i+1<<h-k)-1, mid = l+r>>1;
		int div = pw(prod[i]*prod[i]%P,P-2) * i % P;
		workl(l,mid,div,true);
		ans = (ans + workr(mid+1,r)) % P;
		workl(l,mid,div,false);
	}
	cout<<ans;
	return 0;
}