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>
const long long mod = 1e9 + 7;
const int N = 1e5 + 5;
int n, m;
long long seed, vmax, a[N];
struct Seg {
  int l, r;
  long long x;
  bool operator<(const Seg &b) const { return r < b.r; }
  bool operator>(const Seg &b) const { return x < b.x; }
  Seg(int l = 0, int r = 0, long long x = 0) : l(l), r(r), x(x) {}
};
std::set<Seg> s;
std::map<long long, int> t;
void split(int p) {
  auto it = s.lower_bound(Seg(p, p));
  if (it == s.end() || it->l == p) return;
  int l = it->l, r = it->r;
  long long x = it->x;
  s.erase(it);
  s.insert(Seg(l, p - 1, x));
  s.insert(Seg(p, r, x));
}
long long rnd() {
  long long ret = seed;
  seed = (seed * 7 + 13) % mod;
  return ret;
}
void init() {
  for (int i = 1; i <= n; i++) a[i] = (rnd() % vmax) + 1;
}
long long qpow(long long x, long long y, long long mod) {
  long long res = 1;
  x %= mod;
  for (; y; y >>= 1, x = x * x % mod)
    if (y & 1) res = res * x % mod;
  return res;
}
int main() {
  scanf("%d%d%lld%lld", &n, &m, &seed, &vmax);
  init();
  int op, l, r;
  long long x, y;
  for (int i = 1; i <= n; i++) s.insert(Seg(i, i, a[i]));
  while (m--) {
    op = rnd() % 4 + 1;
    l = rnd() % n + 1;
    r = rnd() % n + 1;
    if (l > r) std::swap(l, r);
    split(l);
    split(r + 1);
    t.clear();
    if (op == 3)
      x = rnd() % (r - l + 1) + 1;
    else
      x = rnd() % vmax + 1;
    if (op == 4) y = rnd() % vmax + 1;
    if (op == 1) {
      while (1) {
        auto it = s.lower_bound(Seg(l, l));
        if (it == s.end() || it->l > r) break;
        l = it->r + 1;
        Seg now = *it;
        now.x += x;
        s.erase(it);
        s.insert(now);
      }
    } else if (op == 2) {
      auto it = s.lower_bound(Seg(l, l));
      while (it != s.end() && it->l <= r) s.erase(it++);
      s.insert(Seg(l, r, x));
    } else if (op == 3) {
      auto it = s.lower_bound(Seg(l, l));
      while (it != s.end() && it->l <= r) t[it->x] += it->r - it->l + 1, it++;
      for (auto it = t.begin(); it != t.end(); it++) {
        x -= it->second;
        if (x <= 0) {
          printf("%lld\n", it->first);
          break;
        }
      }
    } else {
      auto it = s.lower_bound(Seg(l, l));
      while (it != s.end() && it->l <= r) t[it->x] += it->r - it->l + 1, it++;
      long long ans = 0;
      for (auto it = t.begin(); it != t.end(); it++)
        ans = (ans + qpow(it->first, x, y) * (it->second) % y) % y;
      printf("%lld\n", ans);
    }
  }
  return 0;
}