Problem: Steve Rogers is fascinated with new vibranium shields S.H.I.E.L.D gave him. They're all uncolored. There are n shields in total, the i-th shield is located at point (xi, yi) of the coordinate plane. It's possible that two or more shields share the same location.

Steve wants to paint all these shields. He paints each shield in either red or blue. Painting a shield in red costs r dollars while painting it in blue costs b dollars.

Additionally, there are m constraints Steve wants to be satisfied. The i-th constraint is provided by three integers ti, li and di:

  * If ti = 1, then the absolute difference between the number of red and blue shields on line x = li should not exceed di. 
  * If ti = 2, then the absolute difference between the number of red and blue shields on line y = li should not exceed di. 



Steve gave you the task of finding the painting that satisfies all the condition and the total cost is minimum.

Input

The first line of the input contains two integers n and m (1 ≤ n, m ≤ 100 000) — the number of shields and the number of constraints respectively.

The second line contains two integers r and b (1 ≤ r, b ≤ 109).

The next n lines contain the shields coordinates. The i-th of these lines contains two integers xi and yi (1 ≤ xi, yi ≤ 109).

The next m lines contain the constrains. The j-th of these lines contains three integers tj, lj and dj (1 ≤ tj ≤ 2, 1 ≤ lj ≤ 109, 0 ≤ dj ≤ n).

Output

If satisfying all the constraints is impossible print -1 in first and only line of the output.

Otherwise, print the minimum total cost in the first line of output. In the second line print a string of length n consisting of letters 'r' and 'b' only. The i-th character should be 'r' if the i-th shield should be painted red in the optimal answer and 'b' if it should be painted blue. The cost of painting shields in these colors should be equal the minimum cost you printed on the first line.

If there exist more than one optimal solution, print any of them.

Examples

Input

5 6
8 3
2 10
1 5
9 10
9 10
2 8
1 9 1
1 2 1
2 10 3
2 10 2
1 1 1
2 5 2


Output

25
rbrbb


Input

4 4
7 3
10 3
9 8
10 3
2 8
2 8 0
2 8 0
1 2 0
1 9 0


Output

-1

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
namespace {
using namespace std;
template <typename flow_t = long long>
struct PushRelabel {
  struct Edge {
    int to, rev;
    flow_t f, c;
  };
  vector<vector<Edge>> g;
  vector<flow_t> ec;
  vector<Edge*> cur;
  vector<vector<int>> hs;
  vector<int> H;
  PushRelabel(int n) : g(n), ec(n), cur(n), hs(2 * n), H(n) {}
  void add_edge(int s, int t, flow_t cap, flow_t rcap = 0) {
    if (s == t) return;
    Edge a = {t, (int)g[t].size(), 0, cap};
    Edge b = {s, (int)g[s].size(), 0, rcap};
    g[s].push_back(a);
    g[t].push_back(b);
  }
  void add_flow(Edge& e, flow_t f) {
    Edge& back = g[e.to][e.rev];
    if (!ec[e.to] && f) hs[H[e.to]].push_back(e.to);
    e.f += f;
    e.c -= f;
    ec[e.to] += f;
    back.f -= f;
    back.c += f;
    ec[back.to] -= f;
  }
  flow_t max_flow(int s, int t) {
    const int n = g.size();
    ec.assign(n, 0);
    cur.assign(n, nullptr);
    hs.assign(2 * n, {});
    H.assign(n, 0);
    int v = g.size();
    H[s] = v;
    ec[t] = 1;
    vector<int> co(2 * v);
    co[0] = v - 1;
    for (int i = 0; i < v; ++i) cur[i] = g[i].data();
    for (auto& e : g[s]) add_flow(e, e.c);
    if (hs[0].size())
      for (int hi = 0; hi >= 0;) {
        int u = hs[hi].back();
        hs[hi].pop_back();
        while (ec[u] > 0)
          if (cur[u] == g[u].data() + g[u].size()) {
            H[u] = 1e9;
            for (auto& e : g[u])
              if (e.c && H[u] > H[e.to] + 1) H[u] = H[e.to] + 1, cur[u] = &e;
            if (++co[H[u]], !--co[hi] && hi < v)
              for (int i = 0; i < v; ++i)
                if (hi < H[i] && H[i] < v) {
                  --co[H[i]];
                  H[i] = v + 1;
                }
            hi = H[u];
          } else if (cur[u]->c && H[u] == H[cur[u]->to] + 1)
            add_flow(*cur[u], min(ec[u], cur[u]->c));
          else
            ++cur[u];
        while (hi >= 0 && hs[hi].empty()) --hi;
      }
    return -ec[s];
  }
};
template <typename flow_t = long long>
struct MaxFlow {
  static constexpr flow_t INF = numeric_limits<flow_t>::max();
  PushRelabel<flow_t> flow;
  MaxFlow(const int n) : flow(n) {}
  inline void add_arc(const int u, const int v, const flow_t capacity) {
    flow.add_edge(u, v, capacity);
  }
  inline void add_edge(const int u, const int v, const flow_t capacity) {
    flow.add_edge(u, v, capacity, capacity);
  }
  inline void add_infinite_arc(const int u, const int v) {
    flow.add_edge(u, v, INF);
  }
  inline void add_infinite_edge(const int u, const int v) {
    flow.add_edge(u, v, INF, INF);
  }
  flow_t solve(const int s, const int t) { return flow.max_flow(s, t); }
};
template <typename flow_t>
class MaxFlowLR {
 private:
  static constexpr flow_t INF = numeric_limits<flow_t>::max();
  int n;
  vector<flow_t> from_source, to_sink, outgoing_capacity;
  vector<tuple<int, int, flow_t>> demands;
  flow_t total_demand;
  bool invalid;
  PushRelabel<flow_t> flow;

 public:
  MaxFlowLR(const int n)
      : n(n),
        from_source(n),
        to_sink(n),
        outgoing_capacity(n),
        total_demand(0),
        invalid(false),
        flow(n + 2) {}
  void add_arc(const int u, const int v, const flow_t demand,
               const flow_t capacity) {
    if (demand > capacity) {
      invalid = true;
      return;
    }
    if (u != v) {
      from_source[v] += demand;
      to_sink[u] += demand;
      flow.add_edge(u, v, capacity - demand);
      demands.push_back({u, v, demand});
      total_demand += demand;
      outgoing_capacity[u] += capacity;
    }
  }
  flow_t solve(const int s, const int t) {
    if (invalid) {
      return -1;
    }
    const int source = n;
    const int sink = n + 1;
    for (int v = 0; v < n; ++v) {
      if (from_source[v] != 0) {
        flow.add_edge(source, v, from_source[v]);
      }
      if (to_sink[v] != 0) {
        flow.add_edge(v, sink, to_sink[v]);
      }
    }
    flow.add_edge(t, s, INF);
    const flow_t flow_aux = flow.max_flow(source, sink);
    if (s != t) {
      flow.g[s].pop_back();
      flow.g[t].pop_back();
    }
    for (int v = n - 1; v >= 0; --v) {
      if (to_sink[v] != 0) {
        flow.g[sink].pop_back();
        flow.g[v].pop_back();
      }
      if (from_source[v] != 0) {
        flow.g[source].pop_back();
        flow.g[v].pop_back();
      }
    }
    if (flow_aux != total_demand) return -1;
    flow.max_flow(s, t);
    flow_t total_flow = outgoing_capacity[s];
    for (const auto& e : flow.g[s]) total_flow -= e.c;
    return total_flow;
  }
  vector<map<int, flow_t>> retrieve() {
    vector<map<int, flow_t>> ans(n);
    for (int u = 0; u < n; ++u)
      for (const auto& e : flow.g[u])
        if (e.f > 0) ans[u][e.to] += e.f, ans[e.to][u] -= e.f;
    for (auto t : demands)
      if (get<2>(t) > 0)
        ans[get<0>(t)][get<1>(t)] += get<2>(t),
            ans[get<1>(t)][get<0>(t)] -= get<2>(t);
    return ans;
  }
};
}  // namespace
inline int divceil(const int x, const int y) {
  const int q = x / y;
  return q * y == x ? q : q + 1;
}
inline long long divceil(const long long x, const long long y) {
  const long long q = x / y;
  return q * y == x ? q : q + 1;
}
map<int, int> coordinate_compression(vector<int> v) {
  sort(v.begin(), v.end());
  const int n = unique(v.begin(), v.end()) - v.begin();
  map<int, int> compr;
  for (int i = 0; i < n; ++i) compr[v[i]] = i;
  return compr;
}
map<long long, int> coordinate_compression(vector<long long> v) {
  sort(v.begin(), v.end());
  const int n = unique(v.begin(), v.end()) - v.begin();
  map<long long, int> compr;
  for (int i = 0; i < n; ++i) compr[v[i]] = i;
  return compr;
}
int main() {
  int n, m;
  scanf("%d%d", &n, &m);
  int r, b;
  scanf("%d%d", &r, &b);
  bool flip = false;
  if (r >= b) {
    flip = true;
    swap(r, b);
  }
  vector<std::pair<int, int>> original_points;
  vector<int> allx, ally;
  for (int i = 0; i < n; ++i) {
    int x, y;
    scanf("%d%d", &x, &y);
    original_points.emplace_back(x, y);
    allx.push_back(x);
    ally.push_back(y);
  }
  map<int, int> comprx = coordinate_compression(allx),
                compry = coordinate_compression(ally);
  const int xlines = comprx.rbegin()->second + 1;
  const int ylines = compry.rbegin()->second + 1;
  vector<int> xdeg(xlines), ydeg(ylines);
  vector<int> xcons(xlines), ycons(ylines);
  vector<std::pair<int, int>> points;
  for (const std::pair<int, int> p : original_points) {
    const int xx = comprx[p.first];
    const int yy = compry[p.second];
    points.emplace_back(xx, yy);
    ++xdeg[xx];
    ++ydeg[yy];
    ++xcons[xx];
    ++ycons[yy];
  }
  while (m--) {
    int t, l, d;
    scanf("%d%d%d", &t, &l, &d);
    if (t == 1) {
      const auto it = comprx.find(l);
      if (it != comprx.end()) xcons[it->second] = min(xcons[it->second], d);
    } else {
      const auto it = compry.find(l);
      if (it != compry.end()) ycons[it->second] = min(ycons[it->second], d);
    }
  }
  MaxFlowLR<long long> dinic(2 + xlines + ylines);
  const int source = 0;
  const int sink = 1;
  const auto get_xvert = [&](const int i) { return 2 + i; };
  const auto get_yvert = [&](const int i) { return 2 + xlines + i; };
  for (int i = 0; i < xlines; ++i) {
    const int L = divceil(xdeg[i] - xcons[i], 2);
    const int R = (xdeg[i] + xcons[i]) / 2;
    dinic.add_arc(source, get_xvert(i), L, R);
  }
  for (int i = 0; i < ylines; ++i) {
    const int L = divceil(ydeg[i] - ycons[i], 2);
    const int R = (ydeg[i] + ycons[i]) / 2;
    dinic.add_arc(get_yvert(i), sink, L, R);
  }
  for (const std::pair<int, int> p : points)
    dinic.add_arc(get_xvert(p.first), get_yvert(p.second), 0, 1);
  long long flow_value = dinic.solve(source, sink);
  if (flow_value == -1) {
    puts("-1");
    return 0;
  }
  long long ans = flow_value * r + (n - flow_value) * b;
  cout << ans << endl;
  map<std::pair<int, int>, int> red;
  auto flow = dinic.retrieve();
  for (int i = 0; i < xlines; ++i) {
    const int x = get_xvert(i);
    for (const auto p : flow[x]) {
      const int j = p.first - 2 - xlines;
      const int f = p.second;
      if (j >= 0) red[std::pair<int, int>(i, j)] += f;
    }
  }
  for (const std::pair<int, int> p : points) {
    int& v = red[p];
    if (v) {
      v--;
      putchar(flip ? 'b' : 'r');
    } else {
      putchar(flip ? 'r' : 'b');
    }
  }
  cout << endl;
}