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 rep(i, a, b) for(int i = a; i < b; ++i)

const int N = 3e5;
int n, u, v, k, gk, gu, gv, ge, f[30], fe[N], sz[N];
vector <array <int, 2>> g[N];

void dfs(int v, int p, int pe) {
    sz[v] = 1;
    for (auto [u, i] : g[v]) if (u != p && !fe[i])
        dfs(u, v, i), sz[v] += sz[u];
    if (sz[v] == f[gk] || sz[v] == f[gk-1]) gv = v, gu = p, ge = pe;
}
bool check(int root, int k) {
    gk = --k, gv = ge = -1;
    dfs(root, -1, -1);
    if (sz[root] == 1) return 1;
    if (sz[root] != f[k+1] || gv == -1) return 0;
    fe[ge] = 1;
    int u = gv, v = gu;
    if (sz[u] != f[k]) swap(u, v);
    return check(u, k) && check(v, k-1);
}
int main() {
    f[0] = f[1] = 1;
    rep(i, 2, 30) f[i] = f[i - 1] + f[i - 2];
    cin >> n;
    rep(i, 1, n) {
        cin >> u >> v;
        g[u].push_back({v, i});
        g[v].push_back({u, i});
    }
    while (f[k] < n) k++;
    puts((f[k] == n && check(1, k)) ? "YES" : "NO");
}