Problem: Let F_k denote the k-th term of Fibonacci sequence, defined as below:

  * F_0 = F_1 = 1
  * for any integer n ≥ 0, F_{n+2} = F_{n+1} + F_n



You are given a tree with n vertices. Recall that a tree is a connected undirected graph without cycles.

We call a tree a Fib-tree, if its number of vertices equals F_k for some k, and at least one of the following conditions holds:

  * The tree consists of only 1 vertex;
  * You can divide it into two Fib-trees by removing some edge of the tree. 



Determine whether the given tree is a Fib-tree or not.

Input

The first line of the input contains a single integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the number of vertices in the tree.

Then n-1 lines follow, each of which contains two integers u and v (1≤ u,v ≤ n, u ≠ v), representing an edge between vertices u and v. It's guaranteed that given edges form a tree.

Output

Print "YES" if the given tree is a Fib-tree, or "NO" otherwise.

You can print your answer in any case. For example, if the answer is "YES", then the output "Yes" or "yeS" will also be considered as correct answer.

Examples

Input


3
1 2
2 3


Output


YES


Input


5
1 2
1 3
1 4
1 5


Output


NO


Input


5
1 3
1 2
4 5
3 4


Output


YES

Note

In the first sample, we can cut the edge (1, 2), and the tree will be split into 2 trees of sizes 1 and 2 correspondently. Any tree of size 2 is a Fib-tree, as it can be split into 2 trees of size 1.

In the second sample, no matter what edge we cut, the tree will be split into 2 trees of sizes 1 and 4. As 4 isn't F_k for any k, it's not Fib-tree.

In the third sample, here is one possible order of cutting the edges so that all the trees in the process are Fib-trees: (1, 3), (1, 2), (4, 5), (3, 4). 

Solution (in PYTHON3):

Solution:
import sys,random,bisect
from collections import deque,defaultdict
from heapq import heapify,heappop,heappush
from itertools import permutations
from math import log,gcd

input = lambda :sys.stdin.buffer.readline()
mi = lambda :map(int,input().split())
li = lambda :list(mi())

fibo = [1,1]
for i in range(40):
    fibo.append(fibo[-1]+fibo[-2])

fibo_dic = {e:i for i,e in enumerate(fibo)}

N = int(input())
edge = [[] for i in range(N)]
for i in range(N-1):
    u,v = mi()
    edge[u-1].append(v-1)
    edge[v-1].append(u-1)

deq = deque([0])
res = []
size = [0 for i in range(N)]
parent = [-1 for i in range(N)]
while deq:
    v = deq.popleft()
    res.append(v)
    size[v] += 1
    for nv in edge[v]:
        if size[nv]!=1:
            parent[nv] = v
            deq.append(nv)

res = res[::-1]

for v in res:
    for nv in edge[v]:
        if nv!=parent[v]:
            size[v] += size[nv]
    if v:
        edge[v].remove(parent[v])
    edge[v] = [nv for nv in edge[v] if nv!=parent[v]]

if N not in fibo_dic:
    exit(print("NO"))

k = fibo_dic[N]
Nodes = [(0,k)]
cutted = set()

while Nodes:
    root,k = Nodes.pop()
    if size[root]<=2:
        continue

    cut = (-1,-1)
    stack = [root]
    while stack:
        v = stack.pop()
        for nv in edge[v]:
            if size[nv] == fibo[k-1] or size[nv] == fibo[k-2]:
                cut = (v,nv)
                break
            stack.append(nv)
        else:
            continue
        break

    if cut==(-1,-1):
        exit(print("NO"))

    pv,v = cut
    edge[pv].remove(v)
    pos = pv
    while True:
        size[pos] -= size[v]
        if parent[pos]!=-1 and (parent[pos],pos) not in cutted:
            pos = parent[pos]
        else:
            break

    cutted.add(cut)

    Nodes.append((root,fibo_dic[size[root]]))
    Nodes.append((v,fibo_dic[size[v]]))

print("YES")