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 ODT {
  int l, r;
  mutable long long val;
  ODT(int L, int R = -1, long long V = 0) : l(L), r(R), val(V) {}
  bool operator<(const ODT& o) const { return l < o.l; }
};
set<ODT> s;
int n, m;
long long seed, vmax, a[200000];
long long getin() {
  long long res = seed;
  seed = (seed * 7 + 13) % 1000000007;
  return res;
}
set<ODT>::iterator split(int pos) {
  set<ODT>::iterator iter = s.lower_bound(ODT(pos));
  if (iter != s.end() && iter->l == pos) return iter;
  --iter;
  int L = iter->l, R = iter->r;
  long long Val = iter->val;
  s.erase(iter);
  s.insert(ODT(L, pos - 1, Val));
  return s.insert(ODT(pos, R, Val)).first;
}
void add(int l, int r, long long Val) {
  set<ODT>::iterator iterr = split(r + 1), iterl = split(l);
  for (; iterl != iterr; ++iterl) iterl->val += Val;
}
void change(int l, int r, long long Val) {
  set<ODT>::iterator iterr = split(r + 1), iterl = split(l);
  s.erase(iterl, iterr);
  s.insert(ODT(l, r, Val));
}
long long kth(int l, int r, int k) {
  vector<pair<long long, int> > vec;
  set<ODT>::iterator iterr = split(r + 1), iterl = split(l);
  vec.clear();
  for (; iterl != iterr; iterl++)
    vec.push_back(pair<long long, int>(iterl->val, iterl->r - iterl->l + 1));
  sort(vec.begin(), vec.end());
  for (vector<pair<long long, int> >::iterator iter = vec.begin();
       iter != vec.end(); iter++) {
    k -= iter->second;
    if (k <= 0) return iter->first;
  }
}
long long quickpow(long long aa, long long bb, long long mo) {
  long long res = 1;
  long long ans = aa % mo;
  while (bb) {
    if (bb & 1) res = res * ans % mo;
    ans = ans * ans % mo;
    bb >>= 1;
  }
  return res;
}
long long getsum(int l, int r, int ex, int mo) {
  set<ODT>::iterator iterr = split(r + 1), iterl = split(l);
  long long res = 0;
  for (; iterl != iterr; iterl++)
    res = (res + (long long)(iterl->r - iterl->l + 1) *
                     quickpow(iterl->val, (long long)ex, (long long)mo)) %
          mo;
  return res;
}
int main() {
  scanf("%d%d%lld%lld", &n, &m, &seed, &vmax);
  for (int i = 1; i <= n; i++) {
    a[i] = (getin() % vmax) + 1;
    s.insert(ODT(i, i, a[i]));
  }
  s.insert(ODT(n + 1, n + 1, 0));
  for (int i = 1; i <= m; i++) {
    int op = (int)(getin() % 4) + 1;
    int l = (int)(getin() % n) + 1;
    int r = (int)(getin() % n) + 1;
    if (l > r) swap(l, r);
    int xx, yy;
    if (op == 3)
      xx = (int)(getin() % (r - l + 1)) + 1;
    else
      xx = (int)(getin() % vmax) + 1;
    if (op == 4) yy = (int)(getin() % vmax) + 1;
    if (op == 1)
      add(l, r, (long long)xx);
    else if (op == 2)
      change(l, r, (long long)xx);
    else if (op == 3)
      printf("%lld\n", kth(l, r, xx));
    else if (op == 4)
      printf("%lld\n", getsum(l, r, xx, yy));
  }
  return 0;
}