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 CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;

#define ll long long
const int N = 2e5+15;

bool fibonaci[N];
vector<int> adj[N];
int a[N],b[N];
bool ban[N];

int sz[N];
void dfs(int node,int par) {
  sz[node] = 1;
  for(int eid : adj[node]) {
    int ch = a[eid]^b[eid]^node;
    if(ch==par || ban[eid]) continue;
    dfs(ch,node);
    sz[node] += sz[ch];
  }
}

int canCut(int node,int par,int size) {
  for(int eid : adj[node]) {
    int ch = a[eid]^b[eid]^node;
    if(ch==par || ban[eid]) continue;
    if(fibonaci[sz[ch]] && fibonaci[size-sz[ch]]) {
      return eid;
    }
    int cid = canCut(ch,node,size);
    if(cid!=-1) {
      return cid;
    }
  }
  return -1;
}

bool canDo(int node) {
  dfs(node,0);
  int size= sz[node];
  if(!fibonaci[size]) return false;
  if(size<=3) return true;

  int cid = canCut(node,0,size);
  if(cid==-1) {
    return false;
  }
  ban[cid] = true;
  if( canDo(a[cid]) && canDo(b[cid])) return true;
  else return false;
}

void init() {
  fibonaci[1] = fibonaci[2] = 1;
  int x = 1,y = 2;
  while( y<N) {
    fibonaci[y] = true;
    int g = x+y;
    x = y,y = g;
  }
}

void solve2() {
  init();
  int n;
  scanf("%d ",&n);
  for(int i=0;i<n-1;++i) {
    scanf("%d %d ", &a[i],&b[i]);
    adj[a[i]].push_back(i),adj[b[i]].push_back(i);
  }
  printf("%s\n", canDo(1) ? "YES\n" : "NO\n");
}

void solve() {
  // int t;
  // cin>>t;
  // while (t--)
    solve2();
}

inline bool exists_test0 (const std::string& name);
inline bool exists_test1 (const std::string& name);
int main() {
#ifdef LOCAL
  string file_name = "input.txt";
  assert(exists_test0(file_name));
  assert(exists_test1(file_name));
  freopen("input.txt","r",stdin);
#endif
  solve();
  return 0;
}

/*
0. Enough array size? Enough array size? Integer overflow?

1. Think TWICE, Code ONCE!
Are there any counterexamples to your algo?

2. Be careful about the BOUNDARIES!
N=1? P=1? Something about 0?

3. Do not make STUPID MISTAKES!
Time complexity? Memory usage? Precision error?

4. Be careful to use wrong variable.
*/
inline bool exists_test0 (const std::string& name) {
    ifstream f(name.c_str());
    return f.good();
}

inline bool exists_test1 (const std::string& name) {
    if (FILE *file = fopen(name.c_str(), "r")) {
        fclose(file);
        return true;
    } else {
        return false;
    }
}