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;
const long long MAXN = 1000005;
const long long SQRTN = 1003;
const long long LOGN = 22;
const double PI = acos(-1);
const long long INF = 1e18;
const long long MOD = 1e9 + 7;
const long long FMOD = 998244353;
const double eps = 1e-9;
void __print(long long x) { cerr << x; }
void __print(long x) { cerr << x; }
void __print(unsigned x) { cerr << x; }
void __print(unsigned long x) { cerr << x; }
void __print(unsigned long long x) { cerr << x; }
void __print(float x) { cerr << x; }
void __print(double x) { cerr << x; }
void __print(long double x) { cerr << x; }
void __print(char x) { cerr << '\'' << x << '\''; }
void __print(const char *x) { cerr << '\"' << x << '\"'; }
void __print(const string &x) { cerr << '\"' << x << '\"'; }
void __print(bool x) { cerr << (x ? "true" : "false"); }
template <typename T, typename V>
void __print(const pair<T, V> &x) {
  cerr << '{';
  __print(x.first);
  cerr << ',';
  __print(x.second);
  cerr << '}';
}
template <typename T>
void __print(const T &x) {
  long long f = 0;
  cerr << '{';
  for (auto &i : x) cerr << (f++ ? "," : ""), __print(i);
  cerr << "}";
}
void _print() { cerr << "]\n"; }
template <typename T, typename... V>
void _print(T t, V... v) {
  __print(t);
  if (sizeof...(v)) cerr << ", ";
  _print(v...);
}
mt19937 RNG(chrono::steady_clock::now().time_since_epoch().count());
template <typename T>
T gcd(T a, T b) {
  if (b == 0) return a;
  a %= b;
  return gcd(b, a);
}
template <typename T>
T lcm(T a, T b) {
  return (a * (b / gcd(a, b)));
}
long long add(long long a, long long b, long long c = MOD) {
  long long res = a + b;
  return (res >= c ? res % c : res);
}
long long sub(long long a, long long b, long long c = MOD) {
  long long res;
  if (abs(a - b) < c)
    res = a - b;
  else
    res = (a - b) % c;
  return (res < 0 ? res + c : res);
}
long long mul(long long a, long long b, long long c = MOD) {
  long long res = (long long)a * b;
  return (res >= c ? res % c : res);
}
long long muln(long long a, long long b, long long c = MOD) {
  long long res = (long long)a * b;
  return ((res % c) + c) % c;
}
long long mulmod(long long a, long long b, long long m = MOD) {
  long long q = (long long)(((long double)a * (long double)b) / (long double)m);
  long long r = a * b - q * m;
  if (r > m) r %= m;
  if (r < 0) r += m;
  return r;
}
template <typename T>
T binpow(T e, T n) {
  T x = 1, p = e;
  while (n) {
    if (n & 1) x = x * p;
    p = p * p;
    n >>= 1;
  }
  return x;
}
template <typename T>
T binpow2(T e, T n, T m = MOD) {
  T x = 1, p = e;
  while (n) {
    if (n & 1) x = mul(x, p, m);
    p = mul(p, p, m);
    n >>= 1;
  }
  return x;
}
template <typename T>
T extended_euclid(T a, T b, T &x, T &y) {
  T xx = 0, yy = 1;
  y = 0;
  x = 1;
  while (b) {
    T q = a / b, t = b;
    b = a % b;
    a = t;
    t = xx;
    xx = x - q * xx;
    x = t;
    t = yy;
    yy = y - q * yy;
    y = t;
  }
  return a;
}
template <typename T>
T mod_inverse(T a, T n = MOD) {
  T x, y, z = 0;
  T d = extended_euclid(a, n, x, y);
  return (d > 1 ? -1 : sub(x, z, n));
}
const long long FACSZ = 1e4;
long long fact[FACSZ], ifact[FACSZ];
void precom(long long c = MOD) {
  fact[0] = 1;
  for (long long i = 1; i < FACSZ; i++) fact[i] = mul(fact[i - 1], i, c);
  ifact[FACSZ - 1] = mod_inverse(fact[FACSZ - 1], c);
  for (long long i = FACSZ - 1 - 1; i >= 0; i--) {
    ifact[i] = mul(i + 1, ifact[i + 1], c);
  }
}
vector<long long> primes;
void prime_precom() {
  primes.push_back(2);
  for (long long x = 3; primes.size() <= 1e5 + 1; x += 2) {
    bool isPrime = true;
    for (auto p : primes) {
      if (x % p == 0) {
        isPrime = false;
        break;
      }
      if (p * p > x) {
        break;
      }
    }
    if (isPrime) {
      primes.push_back(x);
    }
  }
}
long long ncr(long long n, long long k) {
  if (n < k) return 0;
  if (k == 0) return 1;
  long long res = 1;
  if (k > n - k) k = n - k;
  for (long long i = 0; i < k; ++i) {
    res *= (n - i);
    res /= (i + 1);
  }
  return res;
}
long long ncr_modp(long long n, long long k, long long p = MOD) {
  if (n < k) return 0;
  if (k == 0) return 1;
  long long res = 1;
  if (k > n - k) k = n - k;
  for (long long i = 0; i < k; ++i) {
    res = mul(res, (n - i), MOD);
    res = mul(res, binpow((i + 1), MOD - 2), MOD);
  }
  return res;
}
vector<long long> factors;
void factorize(long long a) {
  factors.clear();
  for (long long i = 1; i * i <= a; i++) {
    if (a % i == 0) {
      factors.push_back(i);
      factors.push_back(a / i);
    }
  }
  sort(factors.begin(), factors.end());
}
long long ncr_precom(long long n, long long r, long long c = MOD) {
  return mul(mul(ifact[r], ifact[n - r], c), fact[n], c);
}
const long long MOD2 = static_cast<long long>(MOD) * MOD;
struct Matrix {
  vector<vector<long long> > mat;
  long long n_rows, n_cols;
  Matrix() {}
  Matrix(vector<vector<long long> > values)
      : mat(values), n_rows(values.size()), n_cols(values[0].size()) {}
  static Matrix identity_matrix(long long n) {
    vector<vector<long long> > values(n, vector<long long>(n, 0));
    for (long long i = 0; i < n; i++) values[i][i] = 1;
    return values;
  }
  Matrix operator*(const Matrix &other) const {
    long long n = n_rows, m = other.n_cols;
    vector<vector<long long> > result(n_rows, vector<long long>(n_cols, 0));
    for (long long i = 0; i < n; i++)
      for (long long j = 0; j < m; j++) {
        long long tmp = 0;
        for (long long k = 0; k < n_cols; k++) {
          tmp += mat[i][k] * 1ll * other.mat[k][j];
          while (tmp >= MOD2) tmp -= MOD2;
        }
        result[i][j] = tmp % MOD;
      }
    return move(Matrix(move(result)));
  }
  inline bool is_square() const { return n_rows == n_cols; }
};
Matrix pw(Matrix a, long long p) {
  Matrix result = Matrix::identity_matrix(a.n_cols);
  while (p > 0) {
    if (p & 1) result = a * result;
    a = a * a;
    p >>= 1;
  }
  return result;
}
vector<long long> v[MAXN];
long long dp[MAXN][2];
vector<long long> xyz(MAXN);
void dfs(long long node, long long parent) {
  if (xyz[node] == 1) {
    dp[node][0] = 1;
  } else if (xyz[node] == 2) {
    dp[node][1] = 1;
  }
  for (auto child : v[node]) {
    if (child == parent) continue;
    dfs(child, node);
    dp[node][0] += dp[child][0];
    dp[node][1] += dp[child][1];
  }
}
long long cnt[3];
void solvethetestcase() {
  long long n;
  cin >> n;
  for (long long i = 0; i < n; i++) {
    cin >> xyz[i + 1];
    cnt[xyz[i + 1]]++;
  }
  for (long long i = 0; i < n - 1; i++) {
    long long a, b;
    cin >> a >> b;
    v[a].push_back(b);
    v[b].push_back(a);
  }
  dfs(1, -1);
  long long ans = 0;
  for (long long i = 2; i < n + 1; i++) {
    if (dp[i][1] == cnt[2] && dp[i][0] == 0) {
      ans++;
      continue;
    }
    if (dp[i][1] == 0 && dp[i][0] == cnt[1]) {
      ans++;
      continue;
    }
  }
  cout << ans << "\n";
}
signed main() {
  cout << fixed << setprecision(12);
  ;
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long t = 1;
  for (long long testcase = 1; testcase < t + 1; testcase++) {
    solvethetestcase();
  }
}