Problem: You are given an undirected tree of n vertices. 

Some vertices are colored blue, some are colored red and some are uncolored. It is guaranteed that the tree contains at least one red vertex and at least one blue vertex.

You choose an edge and remove it from the tree. Tree falls apart into two connected components. Let's call an edge nice if neither of the resulting components contain vertices of both red and blue colors.

How many nice edges are there in the given tree?

Input

The first line contains a single integer n (2 ≤ n ≤ 3 ⋅ 10^5) — the number of vertices in the tree.

The second line contains n integers a_1, a_2, ..., a_n (0 ≤ a_i ≤ 2) — the colors of the vertices. a_i = 1 means that vertex i is colored red, a_i = 2 means that vertex i is colored blue and a_i = 0 means that vertex i is uncolored.

The i-th of the next n - 1 lines contains two integers v_i and u_i (1 ≤ v_i, u_i ≤ n, v_i ≠ u_i) — the edges of the tree. It is guaranteed that the given edges form a tree. It is guaranteed that the tree contains at least one red vertex and at least one blue vertex.

Output

Print a single integer — the number of nice edges in the given tree.

Examples

Input


5
2 0 0 1 2
1 2
2 3
2 4
2 5


Output


1


Input


5
1 0 0 0 2
1 2
2 3
3 4
4 5


Output


4


Input


3
1 1 2
2 3
1 3


Output


0

Note

Here is the tree from the first example:

<image>

The only nice edge is edge (2, 4). Removing it makes the tree fall apart into components \{4\} and \{1, 2, 3, 5\}. The first component only includes a red vertex and the second component includes blue vertices and uncolored vertices.

Here is the tree from the second example:

<image>

Every edge is nice in it.

Here is the tree from the third example:

<image>

Edge (1, 3) splits the into components \{1\} and \{3, 2\}, the latter one includes both red and blue vertex, thus the edge isn't nice. Edge (2, 3) splits the into components \{1, 3\} and \{2\}, the former one includes both red and blue vertex, thus the edge also isn't nice. So the answer is 0.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
class Debugger {
 public:
  Debugger(const std::string& _separator = ", ")
      : first(true), separator(_separator) {}
  template <typename ObjectType>
  Debugger& operator,(const ObjectType& v) {
    if (!first) cerr << separator;
    cerr << v;
    first = false;
    return *this;
  }
  ~Debugger() { cerr << "\n"; }

 private:
  bool first;
  string separator;
};
template <typename T1, typename T2>
inline ostream& operator<<(std::ostream& os, const std::pair<T1, T2>& p) {
  return os << "(" << p.first << ", " << p.second << ")";
}
template <typename T>
inline ostream& operator<<(ostream& os, const vector<T>& v) {
  bool first = true;
  os << "[";
  for (unsigned long long i = 0; i < v.size(); i++) {
    if (!first) os << ", ";
    os << v[i];
    first = false;
  }
  return os << "]";
}
template <typename T>
inline ostream& operator<<(ostream& os, const set<T>& v) {
  bool first = true;
  os << "[";
  for (typename set<T>::const_iterator ii = v.begin(); ii != v.end(); ++ii) {
    if (!first) os << ", ";
    os << *ii;
    first = false;
  }
  return os << "]";
}
template <typename T1, typename T2>
inline ostream& operator<<(ostream& os, const map<T1, T2>& v) {
  bool first = true;
  os << "[";
  for (typename map<T1, T2>::const_iterator ii = v.begin(); ii != v.end();
       ++ii) {
    if (!first) os << ", ";
    os << *ii;
    first = false;
  }
  return os << "]";
}
long long n;
long long massiv[300010];
vector<long long> adj[300010];
long long x, y;
long long used[300010];
long long dp[300010][4];
long long dpp[300010];
pair<long long, long long> dfs(long long v) {
  used[v] = 1;
  for (auto to : adj[v]) {
    if (used[to] == 1) continue;
    pair<long long, long long> child = dfs(to);
    dp[v][1] += child.first;
    dp[v][2] += child.second;
  }
  if (massiv[v] == 1) {
    dp[v][1]++;
    pair<long long, long long> nuj;
    nuj.first = dp[v][1];
    nuj.second = dp[v][2];
    return nuj;
  } else if (massiv[v] == 2) {
    dp[v][2]++;
    pair<long long, long long> nuj;
    nuj.first = dp[v][1];
    nuj.second = dp[v][2];
    return nuj;
  } else {
    pair<long long, long long> nuj;
    nuj.first = dp[v][1];
    nuj.second = dp[v][2];
    return nuj;
  }
}
long long cnt = 0;
void dfss(long long v, long long skolkox, long long skolkoy) {
  used[v] = 1;
  for (auto to : adj[v]) {
    if (used[to] == 1) continue;
    long long onex = dp[v][1] + skolkox;
    long long oney = dp[v][2] + skolkoy;
    long long twox = dp[to][1];
    long long twoy = dp[to][2];
    onex -= twox;
    oney -= twoy;
    long long sig = 0;
    if (onex > 0 && oney == 0)
      sig = 1;
    else if (oney > 0 && onex == 0)
      sig = 1;
    else if (oney == 0 && onex == 0)
      sig = 1;
    long long sigg = 0;
    if (twox > 0 && twoy == 0)
      sigg = 1;
    else if (twoy > 0 && twox == 0)
      sigg = 1;
    else if (twoy == 0 && twox == 0)
      sigg = 1;
    if (sig == 1 && sigg == 1) cnt++;
    dfss(to, onex, oney);
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  ;
  cin >> n;
  for (long long i = 1; i <= n; i++) cin >> massiv[i];
  for (long long i = 1; i < n; i++) {
    cin >> x >> y;
    adj[x].push_back(y);
    adj[y].push_back(x);
  }
  dfs(1);
  for (long long i = 1; i <= n; i++) used[i] = 0;
  dfss(1, 0, 0);
  cout << cnt << "\n";
  return 0;
}