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 io,os
from collections import deque
def solve(N, edges):
    fib = [1, 1]
    while fib[-1] < N:        fib.append(fib[-2] + fib[-1])
    fibToIndex = {x: i for i, x in enumerate(fib)}
    if N not in fibToIndex:      return "NO"
    graph = [[] for i in range(N)]
    for u, v in edges:graph[u].append(v);graph[v].append(u)
    root = 0;    q = [root];    parent = [-1] * N
    for u in q:
        q += graph[u]
        for v in graph[u]:  parent[v] = u;   graph[v].remove(u)
    children = graph;    subTreeSize = [-1] * N
    for u in q[::-1]:        subTreeSize[u] = 1 + sum(subTreeSize[v] for v in children[u])
    def isFib(root, numNodes):
        if numNodes <= 3:            return True
        i = fibToIndex[numNodes];    F1 = fib[i - 1];   F2 = fib[i - 2];   q = deque([root])
        while q:
            u = q.popleft();     count = subTreeSize[u]
            if count == F1 or count == F2:
                p = parent[u];    parent[u] = -1;     children[p].remove(u)
                while parent[p] != -1:    subTreeSize[p] -= count;      p = parent[p]
                return isFib(u, count) and isFib(p, numNodes - count)
            q += graph[u]
        return False
    if isFib(root, N):        return "YES"
    return "NO"
DEBUG = False
if DEBUG:N = 196418;edges = [(i, i + 1) for i in range(N - 1)];print(solve(N, edges));exit()
if __name__ == "__main__":input = io.BytesIO(os.read(0, os.fstat(0).st_size)).readline;(N,) = [int(x) for x in input().split()];edges = [[int(x) - 1 for x in input().split()] for i in range(N - 1)];print(solve(N, edges))