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;
int n,h,tic = 0,p[1<<20];
const int mod = 1000000007;
long long ans,tick[1<<20],val[1<<20];
inline void AddTag(int now,long long bas) {
	bas *= now;
	bas %= mod;
	if (now >= (1<<h)) {
		now = p[now-(1<<h)+1]+(1<<h)-1;
		while (now > 1) {
			if (tick[now>>1] < tic)
				val[now>>1] = 0,tick[now>>1] = tic;
			bas *= now;
			bas %= mod;
			(val[now>>1] += bas) %= mod;
			now >>= 1;
		//	cout << now << " " << val[now] << endl;
		}
		return ;
	}
	AddTag(now<<1,bas);
	AddTag((now<<1)+1,bas);
}
inline void calcAns(int now,long long bas) {
	bas *= now;
	bas %= mod;
	if (now >= (1<<h)) {
		now = p[now-(1<<h)+1]+(1<<h)-1;
		long long tmp = bas*now%mod,caled = 0;
		while (now > 1) {
			//caled *= now;
			//caled %= mod;
			caled *= now;
			caled %= mod;
			now >>= 1;
			tmp *= now;
			tmp %= mod;
			if (tick[now] == tic) {
				//cout << now << " " << tmp << " " << val[now] << " " <<caled << endl;
				(ans += tmp*(val[now]-caled)) %= mod;
				caled = val[now];
			}
		}
		return ;
	}
	calcAns((now<<1),bas);
	calcAns((now<<1)+1,bas);
}
inline void dfs(int now) {
	if (now >= (1<<h))
		return ;
	tic++;
	AddTag(now<<1,now);
	calcAns((now<<1)+1,1);
	//cout << ans << endl;
	dfs(now<<1);
	dfs((now<<1)+1);
}
int main() {
	scanf("%d",&h);
	h--;
	int n = 1<<h;
	for (int i = 1;i <= n; i++)
		scanf("%d",&p[i]);
	dfs(1);
	ans += mod;
	ans %= mod;
	printf("%lld\n",ans);
	return 0;
}
/*
5
6 14 15 7 12 16 5 4 11 9 3 10 8 2 13 1
4
1 2 3 4 5 6 7 8
*/