#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using ull = unsigned long long;
using ld = long double;
using pii = pair<int, int>;
using pll = pair<ll, ll>;
using vi = vector<int>;
using vb = vector<bool>;
using vc = vector<char>;
using vvc = vector<vc>;
using vvi = vector<vi>;
using vvb = vector<vb>;
using vpii = vector<pii>;
using vpll = vector<pll>;
using vll = vector<ll>;
using vvll = vector<vll>;
using vull = vector<ull>;
constexpr int INF = 2000000000;
constexpr ll INFl = 2000000000ll;
constexpr ll INFll = 2000000000000000000ll;
constexpr ll mod = 1e9 + 7;
struct num {
  num(ll val = 0) : v{val} {
    if (v >= mod) v %= mod;
  };
  ll v;
  operator ll() { return v; };
  num operator+(const num& o) const { return {v + o.v}; };
  num operator-(const num& o) const { return {v - o.v + mod}; };
  num operator*(const num& o) const { return {v * o.v}; };
  num operator+=(const num& o) {
    v = num{v + o.v};
    return *this;
  };
  num operator-=(const num& o) {
    v = num{v - o.v + mod};
    return *this;
  };
  num operator*=(const num& o) {
    v = num{v * o.v};
    return *this;
  };
  num operator!() const {
    num R = 1;
    for (__typeof(v + 1) i = (2); i <= (v + 1); ++i) R *= num{i};
    return R;
  }
  num operator^(ll b) const {
    num R{1}, a{v};
    while (b > 0) {
      if (b & 1) R *= a;
      a *= a, b /= 2;
    }
    return R;
  }
  num operator/(const num& o) const { return (*this) * (o ^ (mod - 2)); }
};
template <typename T1, typename T2>
ostream& operator<<(ostream& os, pair<T1, T2> x) {
  os << '(' << x.first << ' ' << x.second << ')';
  return os;
}
template <typename T1, typename T2>
istream& operator>>(istream& is, pair<T1, T2>& x) {
  is >> x.first >> x.second;
  return is;
}
template <typename T>
ostream& operator<<(ostream& os, vector<T>& x) {
  for (T& i : x) os << i << ' ';
  return os;
}
template <typename T>
istream& operator>>(istream& is, vector<T>& x) {
  for (T& i : x) is >> i;
  return is;
}
template <typename T1, typename T2>
pair<T1, T2> operator+(const pair<T1, T2>& A, const pair<T1, T2>& B) {
  return {A.first + B.first, A.second + B.second};
}
template <typename T1, typename T2>
pair<T1, T2> operator-(const pair<T1, T2>& A, const pair<T1, T2>& B) {
  return {A.first - B.first, A.second - B.second};
}
template <typename T1, typename T2>
pair<T1, T2> operator-(const pair<T1, T2>& A) {
  return {-A.first, -A.second};
}
template <typename T1, typename T2>
pair<T1, T2> operator!(const pair<T2, T1>& A) {
  return {A.second, A.first};
}
template <typename T1, typename T2>
bool maxe(T1& A, const T2& B) {
  if (B > A) {
    A = B;
    return 1;
  }
  return 0;
}
template <typename T1, typename T2>
bool mine(T1& A, const T2& B) {
  if (B < A) {
    A = B;
    return 1;
  }
  return 0;
}
struct vertex {
  vi e;
  bool vis = 0;
  bool type = 0;
  int sp = -1;
  int R = -1;
};
int n, m;
int n1, n2, n3;
vector<vertex> G;
bool in() {
  cin >> n >> m >> n1 >> n2 >> n3;
  G.resize(n);
  for (__typeof(m) i = (1); i <= (m); ++i) {
    int a, b;
    cin >> a >> b;
    --a;
    --b;
    G[a].e.emplace_back(b);
    G[b].e.emplace_back(a);
  }
  return true;
}
pii dfs(int v, bool l, int sp) {
  G[v].vis = 1;
  G[v].type = l;
  G[v].sp = sp;
  pii R{1, 0};
  for (auto u : G[v].e) {
    if (!G[u].vis)
      R = R + !dfs(u, !l, sp);
    else if (G[u].type == G[v].type) {
      cout << "NO" << '\n';
      exit(0);
    }
  }
  return R;
}
vpii T{pii{0, 0}};
vvi DP;
vi R;
ll solve() {
  for (__typeof(n - 1) v = (0); v <= (n - 1); ++v)
    if (!G[v].vis) T.emplace_back(dfs(v, 0, T.size()));
  int N = T.size() - 1;
  R.resize(N + 1, -1);
  DP.resize(N + 1, vi(n2 + 1, -1));
  DP[0][0] = 0;
  for (__typeof(N) i = (1); i <= (N); ++i) {
    auto [x, y] = T[i];
    for (__typeof(n2 - x) j = (0); j <= (n2 - x); ++j)
      if (DP[i - 1][j] >= 0) DP[i][j + x] = 0;
    for (__typeof(n2 - y) j = (0); j <= (n2 - y); ++j)
      if (DP[i - 1][j] >= 0) DP[i][j + y] = 1;
  }
  if (DP[N][n2] < 0) return -'n';
  cout << "YES" << '\n';
  int j = n2;
  for (__typeof(1) i = (N); i >= (1); --i) {
    R[i] = DP[i][j];
    if (DP[i][j] == 0)
      j -= T[i].first;
    else if (DP[i][j] == 1)
      j -= T[i].second;
    else
      return -'e';
  }
  for (__typeof(n - 1) v = (0); v <= (n - 1); ++v)
    if (int(G[v].type) == R[G[v].sp])
      G[v].R = 2;
    else
      G[v].R = 1;
  for (__typeof(n - 1) v = (0); v <= (n - 1); ++v)
    if (G[v].R == 1 && n3 > 0) {
      G[v].R = 3;
      n3--;
    }
  for (__typeof(n - 1) v = (0); v <= (n - 1); ++v) cout << G[v].R;
  cout << '\n';
  return -INF;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int noOfTests = 1;
  while (noOfTests-- > 0) {
    if (!in()) {
      cout << "popsute_wejscie\n";
      continue;
    }
    ll result = solve();
    if (result == ll(-'t'))
      cout << "YES\n";
    else if (result == ll(-'n'))
      cout << "NO\n";
    else if (result == ll(-'e'))
      cout << "Popsute\n";
    else if (result != -INF)
      cout << result << '\n';
  }
  return 0;
}

### Prompt

Your task is to create a Cpp solution to the following problem:

You are given an undirected graph without self-loops or multiple edges which consists of n vertices and m edges. Also you are given three integers n_1, n_2 and n_3.

Can you label each vertex with one of three numbers 1, 2 or 3 in such way, that: 

  1. Each vertex should be labeled by exactly one number 1, 2 or 3; 
  2. The total number of vertices with label 1 should be equal to n_1; 
  3. The total number of vertices with label 2 should be equal to n_2; 
  4. The total number of vertices with label 3 should be equal to n_3; 
  5. |col_u - col_v| = 1 for each edge (u, v), where col_x is the label of vertex x. 



If there are multiple valid labelings, print any of them.

Input

The first line contains two integers n and m (1 ≤ n ≤ 5000; 0 ≤ m ≤ 10^5) — the number of vertices and edges in the graph.

The second line contains three integers n_1, n_2 and n_3 (0 ≤ n_1, n_2, n_3 ≤ n) — the number of labels 1, 2 and 3, respectively. It's guaranteed that n_1 + n_2 + n_3 = n.

Next m lines contan description of edges: the i-th line contains two integers u_i, v_i (1 ≤ u_i, v_i ≤ n; u_i ≠ v_i) — the vertices the i-th edge connects. It's guaranteed that the graph doesn't contain self-loops or multiple edges.

Output

If valid labeling exists then print "YES" (without quotes) in the first line. In the second line print string of length n consisting of 1, 2 and 3. The i-th letter should be equal to the label of the i-th vertex.

If there is no valid labeling, print "NO" (without quotes).

Examples

Input


6 3
2 2 2
3 1
5 4
2 5


Output


YES
112323


Input


5 9
0 2 3
1 2
1 3
1 5
2 3
2 4
2 5
3 4
3 5
4 5


Output


NO

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using ull = unsigned long long;
using ld = long double;
using pii = pair<int, int>;
using pll = pair<ll, ll>;
using vi = vector<int>;
using vb = vector<bool>;
using vc = vector<char>;
using vvc = vector<vc>;
using vvi = vector<vi>;
using vvb = vector<vb>;
using vpii = vector<pii>;
using vpll = vector<pll>;
using vll = vector<ll>;
using vvll = vector<vll>;
using vull = vector<ull>;
constexpr int INF = 2000000000;
constexpr ll INFl = 2000000000ll;
constexpr ll INFll = 2000000000000000000ll;
constexpr ll mod = 1e9 + 7;
struct num {
  num(ll val = 0) : v{val} {
    if (v >= mod) v %= mod;
  };
  ll v;
  operator ll() { return v; };
  num operator+(const num& o) const { return {v + o.v}; };
  num operator-(const num& o) const { return {v - o.v + mod}; };
  num operator*(const num& o) const { return {v * o.v}; };
  num operator+=(const num& o) {
    v = num{v + o.v};
    return *this;
  };
  num operator-=(const num& o) {
    v = num{v - o.v + mod};
    return *this;
  };
  num operator*=(const num& o) {
    v = num{v * o.v};
    return *this;
  };
  num operator!() const {
    num R = 1;
    for (__typeof(v + 1) i = (2); i <= (v + 1); ++i) R *= num{i};
    return R;
  }
  num operator^(ll b) const {
    num R{1}, a{v};
    while (b > 0) {
      if (b & 1) R *= a;
      a *= a, b /= 2;
    }
    return R;
  }
  num operator/(const num& o) const { return (*this) * (o ^ (mod - 2)); }
};
template <typename T1, typename T2>
ostream& operator<<(ostream& os, pair<T1, T2> x) {
  os << '(' << x.first << ' ' << x.second << ')';
  return os;
}
template <typename T1, typename T2>
istream& operator>>(istream& is, pair<T1, T2>& x) {
  is >> x.first >> x.second;
  return is;
}
template <typename T>
ostream& operator<<(ostream& os, vector<T>& x) {
  for (T& i : x) os << i << ' ';
  return os;
}
template <typename T>
istream& operator>>(istream& is, vector<T>& x) {
  for (T& i : x) is >> i;
  return is;
}
template <typename T1, typename T2>
pair<T1, T2> operator+(const pair<T1, T2>& A, const pair<T1, T2>& B) {
  return {A.first + B.first, A.second + B.second};
}
template <typename T1, typename T2>
pair<T1, T2> operator-(const pair<T1, T2>& A, const pair<T1, T2>& B) {
  return {A.first - B.first, A.second - B.second};
}
template <typename T1, typename T2>
pair<T1, T2> operator-(const pair<T1, T2>& A) {
  return {-A.first, -A.second};
}
template <typename T1, typename T2>
pair<T1, T2> operator!(const pair<T2, T1>& A) {
  return {A.second, A.first};
}
template <typename T1, typename T2>
bool maxe(T1& A, const T2& B) {
  if (B > A) {
    A = B;
    return 1;
  }
  return 0;
}
template <typename T1, typename T2>
bool mine(T1& A, const T2& B) {
  if (B < A) {
    A = B;
    return 1;
  }
  return 0;
}
struct vertex {
  vi e;
  bool vis = 0;
  bool type = 0;
  int sp = -1;
  int R = -1;
};
int n, m;
int n1, n2, n3;
vector<vertex> G;
bool in() {
  cin >> n >> m >> n1 >> n2 >> n3;
  G.resize(n);
  for (__typeof(m) i = (1); i <= (m); ++i) {
    int a, b;
    cin >> a >> b;
    --a;
    --b;
    G[a].e.emplace_back(b);
    G[b].e.emplace_back(a);
  }
  return true;
}
pii dfs(int v, bool l, int sp) {
  G[v].vis = 1;
  G[v].type = l;
  G[v].sp = sp;
  pii R{1, 0};
  for (auto u : G[v].e) {
    if (!G[u].vis)
      R = R + !dfs(u, !l, sp);
    else if (G[u].type == G[v].type) {
      cout << "NO" << '\n';
      exit(0);
    }
  }
  return R;
}
vpii T{pii{0, 0}};
vvi DP;
vi R;
ll solve() {
  for (__typeof(n - 1) v = (0); v <= (n - 1); ++v)
    if (!G[v].vis) T.emplace_back(dfs(v, 0, T.size()));
  int N = T.size() - 1;
  R.resize(N + 1, -1);
  DP.resize(N + 1, vi(n2 + 1, -1));
  DP[0][0] = 0;
  for (__typeof(N) i = (1); i <= (N); ++i) {
    auto [x, y] = T[i];
    for (__typeof(n2 - x) j = (0); j <= (n2 - x); ++j)
      if (DP[i - 1][j] >= 0) DP[i][j + x] = 0;
    for (__typeof(n2 - y) j = (0); j <= (n2 - y); ++j)
      if (DP[i - 1][j] >= 0) DP[i][j + y] = 1;
  }
  if (DP[N][n2] < 0) return -'n';
  cout << "YES" << '\n';
  int j = n2;
  for (__typeof(1) i = (N); i >= (1); --i) {
    R[i] = DP[i][j];
    if (DP[i][j] == 0)
      j -= T[i].first;
    else if (DP[i][j] == 1)
      j -= T[i].second;
    else
      return -'e';
  }
  for (__typeof(n - 1) v = (0); v <= (n - 1); ++v)
    if (int(G[v].type) == R[G[v].sp])
      G[v].R = 2;
    else
      G[v].R = 1;
  for (__typeof(n - 1) v = (0); v <= (n - 1); ++v)
    if (G[v].R == 1 && n3 > 0) {
      G[v].R = 3;
      n3--;
    }
  for (__typeof(n - 1) v = (0); v <= (n - 1); ++v) cout << G[v].R;
  cout << '\n';
  return -INF;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int noOfTests = 1;
  while (noOfTests-- > 0) {
    if (!in()) {
      cout << "popsute_wejscie\n";
      continue;
    }
    ll result = solve();
    if (result == ll(-'t'))
      cout << "YES\n";
    else if (result == ll(-'n'))
      cout << "NO\n";
    else if (result == ll(-'e'))
      cout << "Popsute\n";
    else if (result != -INF)
      cout << result << '\n';
  }
  return 0;
}

```