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;
struct T {
  int l, r;
  mutable long long v;
  bool operator<(T a) const { return l < a.l; }
};
set<T> s;
vector<pair<long long, int>> v;
using it = set<T>::iterator;
long long qp(long long a, int b, int p) {
  int r = 1;
  for (a %= p; b; b >>= 1, a = a * a % p)
    if (b & 1) r = r * a % p;
  return r;
}
it sp(int x) {
  it p = s.lower_bound({x, 0, 0});
  if (p != s.end() && p->l == x) return p;
  --p;
  int l = p->l, r = p->r;
  long long v = p->v;
  return s.erase(p), s.insert({l, x - 1, v}), s.insert({x, r, v}).first;
}
void add(int l, int r, int x) {
  it q = sp(r + 1), p = sp(l);
  for (; p != q; ++p) p->v += x;
}
void cov(int l, int r, int x) {
  it q = sp(r + 1), p = sp(l);
  s.erase(p, q), s.insert({l, r, x});
}
long long kth(int l, int r, int k) {
  it q = sp(r + 1), p = sp(l);
  for (v.clear(); p != q; ++p) v.push_back({p->v, p->r - p->l + 1});
  sort(v.begin(), v.end());
  for (auto o : v)
    if (k -= o.second, k <= 0) return o.first;
}
int sum(int l, int r, int x, int y) {
  it q = sp(r + 1), p = sp(l);
  int w = 0;
  for (; p != q; ++p) w = (w + (p->r - p->l + 1) * qp(p->v, x, y)) % y;
  return w;
}
int main() {
  int n, m, i, o, l, r, x, u, v, w;
  scanf("%d%d%d%d", &n, &m, &u, &v);
  for (i = 1; i <= n; ++i)
    s.insert({i, i, (w = u, u = (u * 7ll + 13) % 1000000007, w) % v + 1});
  while (m--) {
    o = (w = u, u = (u * 7ll + 13) % 1000000007, w) % 4 + 1,
    l = (w = u, u = (u * 7ll + 13) % 1000000007, w) % n + 1,
    r = (w = u, u = (u * 7ll + 13) % 1000000007, w) % n + 1;
    if (l > r) swap(l, r);
    x = (w = u, u = (u * 7ll + 13) % 1000000007, w) %
            (o == 3 ? (r - l + 1) : v) +
        1;
    if (o == 1)
      add(l, r, x);
    else if (o == 2)
      cov(l, r, x);
    else if (o == 3)
      printf("%lld\n", kth(l, r, x));
    else
      printf("%d\n",
             sum(l, r, x, (w = u, u = (u * 7ll + 13) % 1000000007, w) % v + 1));
  }
  return 0;
}