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;
const int N = 200010;
const int M = 27;
int f[M];

int sz[N] , val[N] , val_cnt;
int needed , needed2;

int n;

vector< int > g[N];

int Col[N] , colCnt = 1;

pair< int , int > found;


void DFS(int node,int prnt,int goodCol){
    sz[node] = 1;
    for(int i = 0 ;i < (int)g[node].size();i++){
        if(g[node][i] == prnt || Col[g[node][i]] != goodCol)
            continue;
        DFS(g[node][i] , node , goodCol);
        sz[node] += sz[g[node][i]];
    }
    if(sz[node] == needed){
        found = make_pair(node , prnt);
    }
    else if(sz[node] == needed2){

        found = make_pair(prnt , node);
    }
}

void DFS(int node,int prnt,int newCol,int goodCol){
    Col[node] = newCol;
    for(int i = 0 ;i < (int)g[node].size();i++){
        if(g[node][i] == prnt || Col[g[node][i]] != goodCol)
            continue;
        DFS(g[node][i] , node , newCol , goodCol);
    }
}

bool calc(int node,int cur,int col){
    if(cur <= 3)
        return true;
    found = make_pair(-1 , -1);
    needed = f[cur - 1];
    needed2 = f[cur - 2];
    DFS(node , -1 , col);
    assert(sz[node] == f[cur]);
    if(found.first == -1)
        return false;
    int ncol = colCnt++;
    int a = found.first;
    int b = found.second;
    DFS(found.first , found.second , ncol , col);
    if(!calc(a , cur - 1,ncol ))
        return false;
    if(!calc(b , cur - 2 , col))
        return false;
    return true;
}

int main(){
    f[0] = f[1] = 1;
    for(int i = 2;i < M;i++){
        f[i] = f[i - 1] + f[i - 2];
    }
    scanf("%d",&n);
    for(int u , v , i = 0 ;i < n - 1;i++){
        scanf("%d%d",&u,&v);
        g[u].push_back(v);
        g[v].push_back(u);
    }
    for(int i = 0;i < M;i++){
        if(f[i] == n){
            if(calc(1 , i , 0)){
                puts("YES");
            }
            else{
                puts("NO");
            }
            return 0;
        }
    }
    puts("NO");
    return 0;
}