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>
inline int in() {
  int k = 0;
  char ch = getchar();
  while (ch < '-') ch = getchar();
  while (ch > '-') k = k * 10 + ch - '0', ch = getchar();
  return k;
}
using std::pair;
using std::set;
using std::vector;
struct node {
  int l, r;
  mutable long long v;
  node(int l = 0, int r = -1, long long v = 0) : l(l), r(r), v(v) {}
  inline bool operator<(const node &a) const { return l < a.l; }
};
set<node> cx;
set<node>::iterator split(int pos) {
  set<node>::iterator it = cx.lower_bound(pos);
  if (it != cx.end() && it->l == pos) return it;
  --it;
  node now = *it;
  cx.erase(it);
  cx.insert(node(now.l, pos - 1, now.v));
  return cx.insert(node(pos, now.r, now.v)).first;
}
inline void assign(int l, int r, int x) {
  set<node>::iterator itr = split(r + 1), itl = split(l);
  cx.erase(itl, itr);
  cx.insert(node(l, r, x));
}
void add_val(int l, int r, int x) {
  set<node>::iterator itr = split(r + 1), itl = split(l);
  for (; itl != itr; ++itl) itl->v += x;
}
long long get_kth(int l, int r, int x) {
  set<node>::iterator itr = split(r + 1), itl = split(l);
  vector<pair<long long, int> > Q;
  Q.clear();
  for (; itl != itr; ++itl)
    Q.push_back(std::make_pair(itl->v, itl->r - itl->l + 1));
  std::sort(Q.begin(), Q.end());
  for (vector<pair<long long, int> >::iterator it = Q.begin(); it != Q.end();
       ++it) {
    x -= it->second;
    if (x <= 0) return it->first;
  }
}
int ksm(int a, int k, int YL) {
  int r = 1;
  while (k) {
    if (k & 1) r = 1ll * r * a % YL;
    a = 1ll * a * a % YL, k >>= 1;
  }
  return r;
}
int sum(int l, int r, int cm, int YL) {
  set<node>::iterator itr = split(r + 1), itl = split(l);
  int ans = 0;
  for (; itl != itr; ++itl)
    ans = (1ll * ksm(itl->v % YL, cm, YL) * (itl->r - itl->l + 1) % YL + ans) %
          YL;
  return ans;
}
int main() {
  int n = in(), m = in(), ret, op, l, r, x, y, seed = in(), vmax = in();
  for (int i = 1; i <= n; ++i)
    y = (ret = seed, seed = (1ll * seed * 7 + 13) % 1000000007, ret) % vmax + 1,
    cx.insert(node(i, i, y));
  cx.insert(node(n + 1, n + 1, 0));
  for (int i = 1; i <= m; ++i) {
    op = (ret = seed, seed = (1ll * seed * 7 + 13) % 1000000007, ret) % 4 + 1,
    l = (ret = seed, seed = (1ll * seed * 7 + 13) % 1000000007, ret) % n + 1,
    r = (ret = seed, seed = (1ll * seed * 7 + 13) % 1000000007, ret) % n + 1;
    if (l > r) std::swap(l, r);
    if (op == 3)
      x = (ret = seed, seed = (1ll * seed * 7 + 13) % 1000000007, ret) %
              (r - l + 1) +
          1;
    else
      x = (ret = seed, seed = (1ll * seed * 7 + 13) % 1000000007, ret) % vmax +
          1;
    if (op == 4)
      y = (ret = seed, seed = (1ll * seed * 7 + 13) % 1000000007, ret) % vmax +
          1;
    if (op == 1) add_val(l, r, x);
    if (op == 2) assign(l, r, x);
    if (op == 3) printf("%lld\n", get_kth(l, r, x));
    if (op == 4) printf("%d\n", sum(l, r, x, y));
  }
  return 0;
}