Problem: — Willem...

— What's the matter?

— It seems that there's something wrong with Seniorious...

— I'll have a look...

<image>

Seniorious is made by linking special talismans in particular order.

After over 500 years, the carillon is now in bad condition, so Willem decides to examine it thoroughly.

Seniorious has n pieces of talisman. Willem puts them in a line, the i-th of which is an integer ai.

In order to maintain it, Willem needs to perform m operations.

There are four types of operations:

  * 1 l r x: For each i such that l ≤ i ≤ r, assign ai + x to ai.
  * 2 l r x: For each i such that l ≤ i ≤ r, assign x to ai.
  * 3 l r x: Print the x-th smallest number in the index range [l, r], i.e. the element at the x-th position if all the elements ai such that l ≤ i ≤ r are taken and sorted into an array of non-decreasing integers. It's guaranteed that 1 ≤ x ≤ r - l + 1.
  * 4 l r x y: Print the sum of the x-th power of ai such that l ≤ i ≤ r, modulo y, i.e. <image>.

Input

The only line contains four integers n, m, seed, vmax (1 ≤ n, m ≤ 105, 0 ≤ seed < 109 + 7, 1 ≤ vmax ≤ 109).

The initial values and operations are generated using following pseudo code:
    
    
      
    def rnd():  
      
        ret = seed  
        seed = (seed * 7 + 13) mod 1000000007  
        return ret  
      
    for i = 1 to n:  
      
        a[i] = (rnd() mod vmax) + 1  
      
    for i = 1 to m:  
      
        op = (rnd() mod 4) + 1  
        l = (rnd() mod n) + 1  
        r = (rnd() mod n) + 1  
      
        if (l > r):   
             swap(l, r)  
      
        if (op == 3):  
            x = (rnd() mod (r - l + 1)) + 1  
        else:  
            x = (rnd() mod vmax) + 1  
      
        if (op == 4):  
            y = (rnd() mod vmax) + 1  
      
    

Here op is the type of the operation mentioned in the legend.

Output

For each operation of types 3 or 4, output a line containing the answer.

Examples

Input

10 10 7 9


Output

2
1
0
3


Input

10 10 9 9


Output

1
1
3
3

Note

In the first example, the initial array is {8, 9, 7, 2, 3, 1, 5, 6, 4, 8}.

The operations are:

  * 2 6 7 9
  * 1 3 10 8
  * 4 4 6 2 4
  * 1 4 5 8
  * 2 1 7 1
  * 4 7 9 4 4
  * 1 2 7 9
  * 4 5 8 1 1
  * 2 5 7 5
  * 4 3 10 8 5

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
using uint = unsigned int;
using ll = long long;
using ull = unsigned long long;
constexpr ll TEN(int n) { return (n == 0) ? 1 : 10 * TEN(n - 1); }
template <class T>
using V = vector<T>;
template <class T>
using VV = V<V<T>>;
struct rng {
  struct A {
    int n;
    const bool operator!=(A r) { return n != r.n; }
    A& operator++() {
      n++;
      return *this;
    }
    int operator*() { return n; }
  };
  int l, r;
  rng(int r) : l(0), r(max(0, r)) {}
  rng(int l, int r) : l(l), r(max(l, r)) {}
  A begin() { return A{l}; }
  A end() { return A{r}; }
};
ll pow_mod(ll x, ll n, ll md) {
  x %= md;
  ll r = 1 % md;
  while (n) {
    if (n & 1) r = (r * x) % md;
    x = (x * x) % md;
    n >>= 1;
  }
  return r;
}
ll seed;
int nxt() {
  auto ret = seed;
  seed = (seed * 7 + 13) % (TEN(9) + 7);
  return ret;
}
using R = tuple<int, int, ll>;
int main() {
  cin.tie(0);
  ios::sync_with_stdio(false);
  cout << setprecision(20);
  int n, m;
  ll vmax;
  cin >> n >> m >> seed >> vmax;
  set<R> s;
  for (int i = 0; i < n; i++) {
    ll aa = nxt() % vmax + 1;
    s.insert(R(i, i + 1, aa));
  }
  for (int i = 0; i < m; i++) {
    int op = nxt() % 4 + 1;
    int l = nxt() % n + 1;
    int r = nxt() % n + 1;
    if (l > r) swap(l, r);
    ll x = 0, y = 0;
    if (op == 3) {
      x = nxt() % (r - l + 1) + 1;
    } else {
      x = nxt() % vmax + 1;
    }
    if (op == 4) {
      y = nxt() % vmax + 1;
    }
    l--;
    {
      auto it = s.lower_bound(R(l, -1, -1));
      if (it != s.begin()) {
        it--;
        int a, b;
        ll c;
        tie(a, b, c) = *it;
        it = s.erase(it);
        if (a != l) s.insert(R(a, l, c));
        if (l != b) s.insert(R(l, b, c));
      }
    }
    {
      auto it = s.lower_bound(R(r, -1, -1));
      if (it != s.begin()) {
        it--;
        int a, b;
        ll c;
        tie(a, b, c) = *it;
        it = s.erase(it);
        if (a != r) s.insert(R(a, r, c));
        if (r != b) s.insert(R(r, b, c));
      }
    }
    if (op == 1) {
      auto it = s.lower_bound(R(l, -1, -1));
      while (it != s.end() && get<1>(*it) <= r) {
        auto u = *it;
        it = s.erase(it);
        get<2>(u) += x;
        s.insert(u);
      }
    } else if (op == 2) {
      auto it = s.lower_bound(R(l, -1, -1));
      while (it != s.end() && get<1>(*it) <= r) {
        it = s.erase(it);
      }
      s.insert(R(l, r, x));
    } else if (op == 3) {
      auto it = s.lower_bound(R(l, -1, -1));
      using P = pair<ll, int>;
      V<P> v;
      while (it != s.end() && get<1>(*it) <= r) {
        int a, b;
        ll c;
        tie(a, b, c) = *it;
        v.push_back(P(c, b - a));
        it++;
      }
      sort(begin(v), end(v));
      ll ans = -1;
      x--;
      for (auto p : v) {
        if (x < p.second) {
          ans = p.first;
          break;
        }
        x -= p.second;
      }
      cout << ans << endl;
    } else {
      ll ans = 0;
      auto it = s.lower_bound(R(l, -1, -1));
      while (it != s.end() && get<1>(*it) <= r) {
        int a, b;
        ll c;
        tie(a, b, c) = *it;
        ans += (b - a) * pow_mod(c, x, y);
        ans %= y;
        it++;
      }
      cout << ans << endl;
    }
  }
  return 0;
}