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 = 2e5 + 7;

int n;

int vis[N];

vector<pair<int, int> > g[N];

int sz[N], fa[N];
vector<int> v;

void dfs(int u, int f) {
    fa[u] = f;
    sz[u] = 1;
    v.push_back(u);
    for (auto it: g[u]) {
        if (vis[it.second]) continue;
        if (f == it.first) continue;
        dfs(it.first, u);
        sz[u] += sz[it.first];
    }
}

int fn[2000];

bool work(int u, int all) {
    if (all <= 3) return true;
    int ok = 0;
    for (int i = 2; i <= 30; i++) {
        if (fn[i] == all) {
            ok = i;
            break;
        }
    }
    if (ok == 0) return false;
    v.clear();
    dfs(u, 0);
    int pos = 0;
    for (int i: v) {
        if (sz[i] == fn[ok - 1] || sz[i] == fn[ok - 2]) {
            pos = i;
        }
    }
    if (pos == 0) return false;
    for (auto it: g[pos]) {
        if (it.first == fa[pos]) {
            vis[it.second] = 1;
        }
    }
    if (sz[pos] == fn[ok - 1]) {
        return work(pos, fn[ok - 1]) && work(u, fn[ok - 2]);
    }
    return work(pos, fn[ok - 2]) && work(u, fn[ok - 1]);


}

int main() {
    fn[0] = 1, fn[1] = 1;
    for (int i = 2; i <= 30; i++) {
        fn[i] = fn[i - 1] + fn[i - 2];
    }

    ios::sync_with_stdio(0);
    cin >> n;

    for (int i = 1; i < n; i++) {
        int u, v;
        cin >> u >> v;
        g[u].push_back({v, i});
        g[v].push_back({u, i});
    }
    if (work(1, n)) {
        cout << "Yes\n";
    } else {
        cout << "No\n";
    }
}