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 maxn=2e5+5;
struct edge{
    int v,next;
}E[maxn<<1];
int vis[maxn<<1];
int head[maxn],tot=1;
void addedge(int u,int v){
    E[++tot].v=v;
    E[tot].next=head[u];
    head[u]=tot;
}
int sz[maxn];
int fib[maxn];
void getsz(int u,int fa){
    sz[u]=1;
    for(int i=head[u];i;i=E[i].next){
        int v=E[i].v;
        if(vis[i]||v==fa)continue;
        getsz(v,u);
        sz[u]+=sz[v];
    }
}
void cut(int u,int fa,int k,int &p1,int &p2,int &kk){
    for(int i=head[u];i;i=E[i].next){
        if(p1)return;
        int v=E[i].v;
        if(v==fa||vis[i])continue;
//        printf("u=%d v=%d\n",u,v);
        if(sz[v]==fib[k-1]||sz[v]==fib[k-2]){
            p1=v,p2=u;
            kk=(sz[v]==fib[k-1])?k-1:k-2;
            break;
        }
        cut(v,u,k,p1,p2,kk);
    }
}
bool check(int u,int k){
    if(k<=2)return 1;
    getsz(u,0);
    int p1=0,p2=0,kk=0;
    cut(u,0,k,p1,p2,kk);
    if(!p1){
        return 0;
    }
    for(int i=head[p1];i;i=E[i].next)if(E[i].v==p2)vis[i]=vis[i^1]=1;
    return check(p1,kk)&&check(p2,(kk==k-1)?k-2:k-1);
}
void init(int n){
    fib[0]=fib[1]=1;
    for(int i=2;i<=maxn-5;i++){
        fib[i]=fib[i-1]+fib[i-2];
        if(fib[i]>n)return;
    }
}
int main () {
    int n;
    scanf("%d",&n);
    init(n);
    for(int i=1;i<=n-1;i++){
        int u,v;
        scanf("%d%d",&u,&v);
        addedge(u,v);
        addedge(v,u);
    }
    int k=0;
    while(fib[k]<n)k++;
    if(n==fib[k] && check(1,k)){
        puts("YES");
    }
    else{
        puts("NO");
    }
}
/*
8
1 2
2 3
3 5
3 4
5 6
5 7
6 8
*/