Problem: In a magical land there are n cities conveniently numbered 1, 2, ..., n. Some pairs of these cities are connected by magical colored roads. Magic is unstable, so at any time, new roads may appear between two cities.

Vicky the witch has been tasked with performing deliveries between some pairs of cities. However, Vicky is a beginner, so she can only complete a delivery if she can move from her starting city to her destination city through a double rainbow. A double rainbow is a sequence of cities c_1, c_2, ..., c_k satisfying the following properties:

  * For each i with 1 ≤ i ≤ k - 1, the cities c_i and c_{i + 1} are connected by a road. 
  * For each i with 1 ≤ i ≤ (k - 1)/(2), the roads connecting c_{2i} with c_{2i - 1} and c_{2i + 1} have the same color. 



For example if k = 5, the road between c_1 and c_2 must be the same color as the road between c_2 and c_3, and the road between c_3 and c_4 must be the same color as the road between c_4 and c_5.

Vicky has a list of events in chronological order, where each event is either a delivery she must perform, or appearance of a new road. Help her determine which of her deliveries she will be able to complete.

Input

The first line contains four integers n, m, c, and q (2 ≤ n ≤ 10^5, 1 ≤ m, c, q ≤ 10^5), denoting respectively the number of cities, the number of roads initially present, the number of different colors the roads can take, and the number of events.

Each of the following m lines contains three integers x, y, and z (1 ≤ x, y ≤ n, 1 ≤ z ≤ c), describing that there initially exists a bidirectional road with color z between cities x and y.

Then q lines follow, describing the events. Each event is one of the following two types: 

  1. + x y z (1 ≤ x, y ≤ n, 1 ≤ z ≤ c), meaning a road with color z appears between cities x and y; 
  2. ? x y (1 ≤ x, y ≤ n), meaning you should determine whether Vicky can make a delivery starting at city x and ending at city y. It is guaranteed that x ≠ y. 



It is guaranteed that at any moment, there is at most one road connecting any pair of cities, and that no road connects a city to itself. It is guaranteed that the input contains at least one event of the second type.

Output

For each event of the second type, print a single line containing "Yes" (without quotes) if the delivery can be made, or a single line containing "No" (without quotes) otherwise.

Example

Input


4 3 2 4
1 2 1
2 3 1
3 4 2
? 1 4
? 4 1
+ 3 1 2
? 4 1


Output


Yes
No
Yes

Note

The following picture corresponds to the sample.

<image>

For her first delivery, Vicky can use the sequence 1, 2, 3, 4 which is a double rainbow. However, she cannot complete the second delivery, as she can only reach city 3. After adding the road between cities 1 and 3, she can now complete a delivery from city 4 to city 1 by using the double rainbow 4, 3, 1.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
long long dx[8] = {0, 1, 0, -1, 1, 1, -1, -1};
long long dy[8] = {1, 0, -1, 0, -1, 1, 1, -1};
using namespace std;
class pa3 {
 public:
  long long x;
  long long y, z;
  pa3(long long x = 0, long long y = 0, long long z = 0) : x(x), y(y), z(z) {}
  bool operator<(const pa3 &p) const {
    if (x != p.x) return x < p.x;
    if (y != p.y) return y < p.y;
    return z < p.z;
  }
  bool operator>(const pa3 &p) const {
    if (x != p.x) return x > p.x;
    if (y != p.y) return y > p.y;
    return z > p.z;
  }
  bool operator==(const pa3 &p) const {
    return x == p.x && y == p.y && z == p.z;
  }
  bool operator!=(const pa3 &p) const {
    return !(x == p.x && y == p.y && z == p.z);
  }
};
class pa4 {
 public:
  long long x;
  long long y, z, w;
  pa4(long long x = 0, long long y = 0, long long z = 0, long long w = 0)
      : x(x), y(y), z(z), w(w) {}
  bool operator<(const pa4 &p) const {
    if (x != p.x) return x < p.x;
    if (y != p.y) return y < p.y;
    if (z != p.z) return z < p.z;
    return w < p.w;
  }
  bool operator>(const pa4 &p) const {
    if (x != p.x) return x > p.x;
    if (y != p.y) return y > p.y;
    if (z != p.z) return z > p.z;
    return w > p.w;
  }
  bool operator==(const pa4 &p) const {
    return x == p.x && y == p.y && z == p.z && w == p.w;
  }
};
class pa2 {
 public:
  long long x, y;
  pa2(long long x = 0, long long y = 0) : x(x), y(y) {}
  pa2 operator+(pa2 p) { return pa2(x + p.x, y + p.y); }
  pa2 operator-(pa2 p) { return pa2(x - p.x, y - p.y); }
  bool operator<(const pa2 &p) const { return y != p.y ? y < p.y : x < p.x; }
  bool operator>(const pa2 &p) const { return x != p.x ? x < p.x : y < p.y; }
  bool operator==(const pa2 &p) const {
    return abs(x - p.x) == 0 && abs(y - p.y) == 0;
  }
  bool operator!=(const pa2 &p) const {
    return !(abs(x - p.x) == 0 && abs(y - p.y) == 0);
  }
};
string itos(long long i) {
  ostringstream s;
  s << i;
  return s.str();
}
long long gcd(long long v, long long b) {
  if (v > b) return gcd(b, v);
  if (v == b) return b;
  if (b % v == 0) return v;
  return gcd(v, b % v);
}
long long mod;
long long extgcd(long long a, long long b, long long &x, long long &y) {
  if (b == 0) {
    x = 1;
    y = 0;
    return a;
  }
  long long d = extgcd(b, a % b, y, x);
  y -= a / b * x;
  return d;
}
pair<long long, long long> operator+(const pair<long long, long long> &l,
                                     const pair<long long, long long> &r) {
  return {l.first + r.first, l.second + r.second};
}
pair<long long, long long> operator-(const pair<long long, long long> &l,
                                     const pair<long long, long long> &r) {
  return {l.first - r.first, l.second - r.second};
}
long long pr[200010];
long long inv[200010];
long long beki(long long wa, long long rr, long long warukazu) {
  if (rr == 0) return 1 % warukazu;
  if (rr == 1) return wa % warukazu;
  wa %= warukazu;
  if (rr % 2 == 1)
    return ((long long)beki(wa, rr - 1, warukazu) * (long long)wa) % warukazu;
  long long zx = beki(wa, rr / 2, warukazu);
  return (zx * zx) % warukazu;
}
long long comb(long long nn, long long rr) {
  if (rr < 0 || rr > nn || nn < 0) return 0;
  long long r = pr[nn] * inv[rr];
  r %= mod;
  r *= inv[nn - rr];
  r %= mod;
  return r;
}
void gya(long long ert) {
  pr[0] = 1;
  for (long long i = 1; i <= ert; i++) {
    pr[i] = (pr[i - 1] * i) % mod;
  }
  inv[ert] = beki(pr[ert], mod - 2, mod);
  for (long long i = ert - 1; i >= 0; i--) {
    inv[i] = inv[i + 1] * (i + 1) % mod;
  }
}
set<long long> se[100020];
struct unionfind {
 private:
 public:
  vector<long long> par, ranks, kosuu;
  void shoki(long long N) {
    par.resize(N + 1, 0);
    ranks.resize(N + 1, 0);
    kosuu.resize(N + 1, 1);
    for (long long i = 0; i <= N; i++) {
      par[i] = i;
    }
  }
  long long root(long long x) {
    return par[x] == x ? x : par[x] = root(par[x]);
  }
  bool same(long long x, long long y) { return root(x) == root(y); }
  bool is_root(long long x) { return x == root(x); }
  void unite(long long x, long long y) {
    x = root(x);
    y = root(y);
    long long xx = kosuu[x], yy = kosuu[y];
    if (x == y) return;
    long long oya;
    if (se[x].size() < se[y].size()) {
      for (auto it = se[x].begin(); it != se[x].end(); it++) {
        se[y].insert(*it);
      }
      se[x].clear();
      oya = y;
    } else {
      for (auto it = se[y].begin(); it != se[y].end(); it++) {
        se[x].insert(*it);
      }
      se[y].clear();
      oya = x;
    }
    if (ranks[x] < ranks[y]) {
      par[x] = y;
      kosuu[y] = yy + xx;
      if (oya == x) {
        swap(se[x], se[y]);
      }
    } else {
      par[y] = x;
      if (ranks[x] == ranks[y]) ranks[x] = ranks[x] + 1;
      kosuu[x] = yy + xx;
      if (oya == y) {
        swap(se[x], se[y]);
      }
    }
    return;
  }
};
vector<pair<long long, long long> > G[100020];
vector<pa4> que;
map<long long, long long> ma[100020];
signed main() {
  cin.tie(0);
  ios::sync_with_stdio(false);
  long long n, m, c, q;
  cin >> n >> m >> c >> q;
  unionfind uf;
  uf.shoki(n + 5);
  for (long long i = 0; i < m; i++) {
    long long y, yy, yyy;
    cin >> y >> yy >> yyy;
    que.push_back({0, y, yy, yyy});
  }
  for (long long i = 0; i < q; i++) {
    string s;
    cin >> s;
    if (s == "+") {
      long long y, yy, yyy;
      cin >> y >> yy >> yyy;
      que.push_back({0, y, yy, yyy});
    } else {
      long long y, yy;
      cin >> y >> yy;
      que.push_back({1, y, yy, 0});
    }
  }
  for (auto V : que) {
    if (V.x) {
      long long s = V.y, t = V.z;
      if (uf.same(s, t)) {
        cout << "Yes" << endl;
        continue;
      }
      long long ss = uf.root(s);
      if (se[ss].count(t)) {
        cout << "Yes" << endl;
      } else
        cout << "No" << endl;
    } else {
      long long s = V.y, t = V.z, c = V.w;
      if (ma[s].count(c)) {
        uf.unite(t, ma[s][c]);
      }
      if (ma[t].count(c)) {
        uf.unite(s, ma[t][c]);
      }
      ma[s][c] = t;
      ma[t][c] = s;
      long long ss = uf.root(s);
      long long tt = uf.root(t);
      se[ss].insert(t);
      se[tt].insert(s);
    }
  }
  return 0;
}