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 fi first
#define se second
typedef pair<int,int> pii;
typedef  long long ll;
const int maxn=2e5+5,mod=1e7;
int n;
int head[maxn],cnt;
int fac[maxn];
int sz[maxn];
bool vis[maxn];// 判断是否为斐波那契数
vector<pii > vec;
unordered_map<int,int> mp[maxn]; //判断这条边是否已经被拆了
struct edge{
    int to,next;
}e[maxn<<1];
void add(int u,int v){
    e[++cnt]={v,head[u]};
    head[u]=cnt;
}
void get_size(int son,int fa){// 获得每个节点的子树节点数量
    vec.push_back({son,fa});
    sz[son]=1;
    for(int i=head[son];i;i=e[i].next){
        if(e[i].to==fa||!mp[son][e[i].to]) continue;
        get_size(e[i].to,son);
        sz[son]+=sz[e[i].to];
    }
}
bool check(int son,int fa,int num){
    if(!vis[num]) return 0;
    if(num==1) return 1;
    vec.clear();// vec代表son子树下的所有节点
    get_size(son,fa);
    for(int i=0;i<vec.size();i++){
        int x=vec[i].fi;
        int y=vec[i].se;
        if(x==son) continue;
        int temp=sz[x];// 要用一个变量保存因为sz[x]会改变
        if(vis[temp]&&vis[num-temp]){
            mp[x][y]=mp[y][x]=0;// 代表这条边没了
            return check(x,y,temp)&&check(y,x,num-temp);
        }
    }
    return 0;
}
int main(){
    fac[0]=fac[1]=1;
    vis[1]=1;
    for(int i=2;i<=26;i++){
        fac[i]=fac[i-1]+fac[i-2];
        vis[fac[i]]=1;
    }
    scanf("%d",&n);
    for(int i=1,u,v;i<=n-1;i++){
        scanf("%d%d",&u,&v);
        add(u,v),add(v,u);
        mp[u][v]=mp[v][u]=1;
    }
    printf(check(1,1,n)?"YES\n":"NO\n");
    return 0;
}