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>
/*#ifndef LOCAL_RUN
    #pragma GCC optimize("Ofast")
    #pragma GCC optimize("unroll-loops")
    #pragma GCC optimize("fast-math")
    #pragma GCC target("avx2,tune=native")
#endif*/
#define fr first
#define sc second
#define m_p make_pair
using namespace std;

typedef long long ll;
typedef pair<int,int> pii;

const int N = 2e5 + 10;

vector<int> graph[N];
int fval[64], sz[N], fnum[N], par[N];
bool ban[N];

void dfs(int v, int p = -1)
{
    par[v] = p;
    sz[v] = 1;
    for (int u: graph[v]) {
        if (u != p && !ban[u]) {
            dfs(u, v);
            sz[v] += sz[u];
        }
    }
}

int srch(int v, int n, int p = -1)
{
    if (fnum[sz[v]] != 0 && fnum[n - sz[v]] != 0)
        return v;
    for (int u: graph[v]) {
        if (u == p || ban[u]) continue;
        int x = srch(u, n, v);
        if (x != -1)
            return x;
    }
    return -1;
}

bool solve(int v)
{
    dfs(v);
    if (sz[v] == 1) return true;
    if (!fnum[sz[v]]) return false;
    int x = srch(v, sz[v]), y = par[x];
    if (x == -1)
        return false;
    ban[y] = true;
    if (!solve(x)) return false;
    ban[y] = false;

    ban[x] = true;
    return solve(y);
}

int main()
{
    #ifdef LOCAL_RUN
        freopen("input.txt", "r", stdin);
    #else
        //freopen("vase.in", "r", stdin);
        //freopen("vase.out", "w", stdout);
        ios::sync_with_stdio(false);
        cin.tie(0);
    #endif

    int n;
    cin >> n;
    fnum[fval[1] = 1] = 1;
    int k = 1;
    while (fval[k] < n) {
        ++k;
        fnum[fval[k] = fval[k - 1] + fval[k - 2]] = k;
    }
    for (int i = 1, x, y; i < n; ++i) {
        cin >> x >> y;
        graph[x].push_back(y);
        graph[y].push_back(x);
    }
    cout << (solve(1) ? "YES" : "NO");

    return 0;
}

// 272472LF1