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 <cstdio>
#include <algorithm>
using namespace std;

const int N = (1 << 18);
const int P = (int)(1E9) + 7;

inline int add(int x, int y) {x += y; return x >= P ? x - P : x;}
inline int sub(int x, int y) {x -= y; return x < 0 ? x + P : x;}
inline int mul(int x, int y) {return (int)(1ll * x * y % P);}
int mpow(int b, int p) {
	int r; for(r = 1; p; p >>= 1, b = mul(b, b))
		if( p & 1 ) r = mul(r, b);
	return r;
}

int a[N + 5], b[N + 5];

int ch[2][20*N + 5], cnt[20*N + 5], ncnt;
void insert(int &x, int d, int p, int k) {
	x = (++ncnt), cnt[x] = add(cnt[x], k);
	if( d == 1 ) return ;
	insert(ch[(p >> (d - 2)) & 1][x], d - 1, p, k); 
}

int tmp;
int merge(int x, int y, int p, int d) {
	if( !x || !y ) return x + y;
	tmp = add(tmp, mul(mul(cnt[ch[0][x]], cnt[ch[1][y]]), b[p]));
	tmp = add(tmp, mul(mul(cnt[ch[1][x]], cnt[ch[0][y]]), b[p]));
	
	ch[0][x] = merge(ch[0][x], ch[0][y], p << 1, d - 1);
	ch[1][x] = merge(ch[1][x], ch[1][y], p << 1 | 1, d - 1);	
	cnt[x] = add(cnt[x], cnt[y]); return x;
}

int rt[N + 5];
int main() {
	int h, n, m; scanf("%d", &h), n = (1 << h), m = (n >> 1);
	
	a[0] = b[0] = 1;
	for(int i=1;i<n;i++)
		a[i] = mul(a[i >> 1], i), b[i] = mpow(mul(a[i], a[i >> 1]), P - 2);
		
	for(int i=0,p;i<m;i++)
		scanf("%d", &p), insert(rt[m + i], h, p - 1, mul(a[m + i], a[m + p - 1]));
		
	int ans = 0;
	for(int i=m-1;i>=1;i--) {
		tmp = 0, rt[i] = merge(rt[i << 1], rt[i << 1 | 1], 1, h);
		ans = add(ans, mul(tmp, b[i]));
	}
	printf("%d\n", ans);
}