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>

using namespace std;
const int MAXN = 1 << 18 , p = 1e9 + 7;
int H , L;
int inv[MAXN];
int f1[MAXN] , f2[MAXN];
int ans = 0;
int pos[MAXN];
int tot[MAXN];
void insert(int x , int v){
	while(x > 1){
		v = 1ll * x * v % p;
		tot[x] = (tot[x] + v) % p;
		x >>= 1;
	}
}

void find(int x , int v){
	while(x > 1){
		v = 1ll * x * v % p;
		ans = (ans + 1ll * tot[x ^ 1] * v % p * (x >> 1)) % p;
		x >>= 1;
	}
}

int main(){
	scanf("%d" , &H);
	L = (1 << (H - 1));
	inv[0] = inv[1] = 1;
	for(int i = 2 ; i < MAXN ; ++ i)
		inv[i] = p - 1ll * (p / i) * inv[p % i] % p;
	f1[0] = f1[1] = f2[0] = f2[1] = 1;
	for(int i = 2 ; i < MAXN ; ++ i){
		f1[i] = 1ll * f1[i / 2] * i % p;
		f2[i] = 1ll * f2[i / 2] * inv[i] % p;
	}
	for(int i = 0 ; i < L ; ++ i){
		scanf("%d" , &pos[i]);
		pos[i] --;
	}
	for(int i = 1 ; i < H ; ++ i)
		for(int j = 0 ; j < L ; j += (1 << i)){
			int u = ((j + L) >> i);
			for(int k = j ; k < j + (1 << (i - 1)) ; ++ k)
				insert(pos[k] + L , 1ll * f1[k + L] * f2[u >> 1] % p);
			for(int k = j + (1 << (i - 1)) ; k < j + (1 << i) ; ++ k)
				find(pos[k] + L , 1ll * f1[k + L] * f2[u] % p);
			for(int k = j ; k < j + (1 << (i - 1)) ; ++ k)
				insert(pos[k] + L , p - 1ll * f1[k + L] * f2[u >> 1] % p);
		}
	printf("%d\n" , ans);
	return 0;
}