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
import sys

h, *ppp = map(int, sys.stdin.buffer.read().split())
n = 1 << (h - 1)
ppp = [p + n - 1 for p in ppp]
MOD = 10 ** 9 + 7

cumprods = [1] * (2 * n)  # 根から頂点vまでの経路の累積積
cumprods_rev = [1] * n  # ..の逆数
cumprods_through = [0] * n  # 木1の根から木2の根まで、木1の i 番目の葉を経由する経路の積
for i in range(2, 2 * n):
    cumprods[i] = cumprods[i >> 1] * i % MOD
for i in range(n):
    cumprods_rev[i] = pow(cumprods[i], MOD - 2, MOD)
    cumprods_through[i] = cumprods[i + n] * cumprods[ppp[i]] % MOD

# 現在考慮中の木1側のLCAを起点として、木2側の各頂点に至るまでの累積積
# 木1からたどり着かないものは0
cumprods_from_tree1 = [0] * (2 * n)

ans = 0

for lca in range(1, n):
    digit = h - lca.bit_length()
    leftmost_leaf_in_left_subtree = lca << digit
    leftmost_leaf_in_right_subtree = ((lca << 1) + 1) << (digit - 1)
    rightmost_leaf_in_right_subtree = (lca + 1) << digit

    rev = cumprods_rev[lca >> 1]
    for leaf in range(leftmost_leaf_in_left_subtree, leftmost_leaf_in_right_subtree):
        v = ppp[leaf - n]
        cp = cumprods_through[leaf - n] * rev % MOD
        while v > 1:
            cumprods_from_tree1[v] += cp * cumprods_rev[v >> 1] % MOD
            v >>= 1

    rev = cumprods_rev[lca]
    for leaf in range(leftmost_leaf_in_right_subtree, rightmost_leaf_in_right_subtree):
        v = ppp[leaf - n]
        cp = cumprods_through[leaf - n] * rev % MOD
        while v > 1:
            ans = (ans + cumprods_from_tree1[v ^ 1] * cp * cumprods_rev[v >> 2]) % MOD
            v >>= 1

    for leaf in range(leftmost_leaf_in_left_subtree, leftmost_leaf_in_right_subtree):
        v = ppp[leaf - n]
        while cumprods_from_tree1[v] != 0:
            cumprods_from_tree1[v] = 0
            v >>= 1

print(ans)
