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;
template <typename _T>
inline void read(_T &f) {
  f = 0;
  _T fu = 1;
  char c = getchar();
  while (c < '0' || c > '9') {
    if (c == '-') fu = -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9') {
    f = (f << 3) + (f << 1) + (c & 15);
    c = getchar();
  }
  f *= fu;
}
template <typename T>
void print(T x) {
  if (x < 0) putchar('-'), x = -x;
  if (x < 10)
    putchar(x + 48);
  else
    print(x / 10), putchar(x % 10 + 48);
}
template <typename T>
void print(T x, char t) {
  print(x);
  putchar(t);
}
const int N = 1e5 + 5;
struct node_t {
  int l, r;
  mutable long long v;
  node_t() {}
  node_t(int L, int R, long long V) : l(L), r(R), v(V) {}
  bool operator<(const node_t A) const { return l < A.l; }
};
set<node_t> rt;
int a[N];
int n, m, seed, vmax;
inline int rnd() {
  int ans = seed;
  seed = ((long long)seed * 7 + 13) % 1000000007;
  return ans;
}
set<node_t>::iterator split(int pos) {
  set<node_t>::iterator it = rt.lower_bound(node_t(pos, pos, 0));
  if (it != rt.end() && it->l == pos) return it;
  --it;
  int l = it->l, r = it->r;
  long long v = it->v;
  rt.erase(it);
  rt.insert(node_t(l, pos - 1, v));
  return rt.insert(node_t(pos, r, v)).first;
}
void assign(int l, int r, int v) {
  split(l);
  set<node_t>::iterator itr = split(r + 1), itl = split(l);
  rt.erase(itl, itr);
  rt.insert(node_t(l, r, v));
}
void add(int l, int r, int v) {
  split(l);
  set<node_t>::iterator itr = split(r + 1), itl = split(l);
  for (; itl != itr; ++itl) itl->v += v;
}
vector<pair<long long, int> > wxw;
long long rnk(int l, int r, int k) {
  wxw.clear();
  split(l);
  set<node_t>::iterator itr = split(r + 1), itl = split(l);
  for (; itl != itr; ++itl)
    wxw.push_back(make_pair(itl->v, itl->r - itl->l + 1));
  sort(wxw.begin(), wxw.end());
  for (register unsigned i = 0; i < wxw.size(); i++) {
    k -= wxw[i].second;
    if (k <= 0) return wxw[i].first;
  }
  throw "k > r - l + 1";
}
inline int fpow(int x, int y, int md) {
  int ans = 1;
  while (y) {
    if (y & 1) ans = (long long)ans * x % md;
    y >>= 1;
    x = (long long)x * x % md;
  }
  return ans;
}
long long sum(int l, int r, int k, int md) {
  split(l);
  set<node_t>::iterator itr = split(r + 1), itl = split(l);
  int ans = 0;
  for (; itl != itr; ++itl) {
    ans += (long long)fpow(itl->v % md, k, md) * (itl->r - itl->l + 1) % md;
    if (ans >= md) ans -= md;
  }
  return ans;
}
int main() {
  cin >> n >> m >> seed >> vmax;
  for (register int i = 1; i <= n; i++)
    a[i] = rnd() % vmax + 1, rt.insert(node_t(i, i, a[i]));
  rt.insert(node_t(n + 1, n + 1, 0));
  while (m--) {
    int opt = rnd() % 4 + 1, l = rnd() % n + 1, r = rnd() % n + 1;
    if (l > r) swap(l, r);
    if (opt == 1) {
      int x = rnd() % vmax + 1;
      add(l, r, x);
    }
    if (opt == 2) {
      int x = rnd() % vmax + 1;
      assign(l, r, x);
    }
    if (opt == 3) {
      int k = rnd() % (r - l + 1) + 1;
      print(rnk(l, r, k), '\n');
    }
    if (opt == 4) {
      int x = rnd() % vmax + 1, y = rnd() % vmax + 1;
      print(sum(l, r, x, y), '\n');
    }
  }
  return 0;
}