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>

#define lowbit(x) ((x)&(-(x)))
using namespace std;
const int N = 2e5 + 10;
int head[N], ver[N << 1], Next[N << 1], tot;
int siz[N], dfn[N], fib[27], num, n;
vector<int> V[27];

struct BIT {
    int t[N];

    inline int ask(int x) {
        int ans = 0;
        for (int i = dfn[x] + siz[x] - 1; i; i -= lowbit(i))ans += t[i];
        for (int i = dfn[x] - 1; i; i -= lowbit(i))ans -= t[i];
        return ans;
    }

    inline void add(int x, int v) {
        for (int i = dfn[x]; i <= n; i += lowbit(i))t[i] += v;
    }
} B;

inline void add(int x, int y) {
    ver[++tot] = y;
    Next[tot] = head[x];
    head[x] = tot;
}

void dfs(int x, int f) {
    siz[x] = 1, dfn[x] = ++num;
    for (int i = head[x], y; i; i = Next[i])
        if ((y = ver[i]) != f)
            dfs(y, x), siz[x] += siz[y];
}

bool solve(int u) {
    if (u <= 3)return true;
    int f1 = fib[u - 1], f2 = fib[u - 2];
    for (int i = 0; i < fib[u]; i++) {
        int x = V[u][i], s = B.ask(x);
        if (s == f1 || s == f2) {
            int u1 = s == f1 ? u - 1 : u - 2;
            int u2 = s == f1 ? u - 2 : u - 1;
            V[u1].clear();
            for (int j = 0, y; j < fib[u]; j++) {
                y = V[u][j];
                if (dfn[y] >= dfn[x] && dfn[y] <= dfn[x] + siz[x] - 1)
                    V[u1].push_back(y);
            }
            if (!solve(u1))continue;
            V[u2].clear();
            for (int j = 0, y; j < fib[u]; j++) {
                y = V[u][j];
                if (dfn[y] < dfn[x] || dfn[y] > dfn[x] + siz[x] - 1)
                    V[u2].push_back(y);
            }
            B.add(x, -s);
            bool res = solve(u2);
            B.add(x, s);
            if (res)return true;
        }
    }
    return false;
}

int main() {
    fib[0] = fib[1] = 1;
    for (int i = 2; i <= 26; i++)
        fib[i] = fib[i - 1] + fib[i - 2];
    scanf("%d", &n);
    if (n <= 3)return puts("YES"), 0;
    int u = lower_bound(fib, fib + 27, n) - fib;
    if (fib[u] != n)return puts("NO"), 0;
    for (int i = 1, x, y; i <= n - 1; i++) {
        scanf("%d%d", &x, &y);
        add(x, y), add(y, x);
    }
    dfs(1, 0);
    for (int i = 1; i <= n; i++)B.add(i, 1);
    for (int i = 1; i <= n; i++)V[u].push_back(i);
    puts(solve(u) ? "YES" : "NO");
    return 0;
}