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 fcin(t &x) {
  int sign = 1;
  x = 0;
  char op = getchar();
  while (op < '0' || op > '9') {
    if (op == '-') sign = -1;
    op = getchar();
  }
  while (op >= '0' && op <= '9') {
    x = x * 10 + (op - 48);
    op = getchar();
  }
  x *= sign;
}
struct node {
  int L, R;
  mutable long long v;
  node(int x, int y = -1, long long z = 0) : L(x), R(y), v(z) {}
  bool operator<(const node &x) const { return L < x.L; }
};
set<node> s;
long long n, m, vmax, seed;
template <typename t>
inline t rnd() {
  t ret = (t)seed;
  seed = (seed * 7 + 13) % 1000000007;
  return ret;
}
long long fastpow(long long a, long long b, long long mod) {
  long long res = 1;
  long long ans = a % mod;
  while (b) {
    if (b & 1) res = res * ans % mod;
    ans = ans * ans % mod;
    b >>= 1;
  }
  return res;
}
set<node>::iterator split(int pos) {
  set<node>::iterator iter = s.lower_bound(node(pos));
  if (iter != s.end() && iter->L == pos) return iter;
  iter--;
  int x = iter->L, y = iter->R;
  long long z = iter->v;
  s.erase(iter);
  s.insert(node(x, pos - 1, z));
  return s.insert(node(pos, y, z)).first;
}
inline void updata(int l, int r, long long d) {
  set<node>::iterator iterr = split(r + 1), iterl = split(l);
  while (iterl != iterr) {
    iterl->v += d;
    iterl++;
  }
}
inline void fill(int l, int r, long long d) {
  set<node>::iterator iterr = split(r + 1), iterl = split(l);
  s.erase(iterl, iterr);
  s.insert(node(l, r, d));
}
long long kth(int l, int r, int k) {
  vector<pair<long long, int> > X;
  set<node>::iterator iterr = split(r + 1), iterl = split(l);
  while (iterl != iterr)
    X.push_back(make_pair(iterl->v, iterl->R - iterl->L + 1)), iterl++;
  sort(X.begin(), X.end());
  for (vector<pair<long long, int> >::iterator iter = X.begin();
       iter != X.end(); iter++) {
    k -= iter->second;
    if (k <= 0) return iter->first;
  }
}
long long sum(int l, int r, long long ex, long long mod) {
  set<node>::iterator iterr = split(r + 1), iterl = split(l);
  long long ans = 0;
  while (iterl != iterr)
    ans = (ans + (long long)(iterl->R - iterl->L + 1) *
                     fastpow(iterl->v, ex, mod) % mod) %
          mod,
    iterl++;
  return ans;
}
int main() {
  fcin(n);
  fcin(m);
  fcin(seed);
  fcin(vmax);
  long long x, y, p;
  int l, r, k;
  for (int i = 1; i <= n; i++)
    p = (rnd<long long>() % vmax) + 1, s.insert(node(i, i, p));
  s.insert(node(n + 1, n + 1, 0));
  for (int i = 1; i <= m; i++) {
    p = rnd<int>() % 4 + 1;
    l = rnd<int>() % n + 1;
    r = rnd<int>() % n + 1;
    if (l > r) swap(l, r);
    if (p == 3)
      k = (rnd<int>() % (r - l + 1)) + 1;
    else
      x = (rnd<long long>() % vmax) + 1;
    if (p == 4) y = (rnd<long long>() % vmax) + 1;
    if (p == 1)
      updata(l, r, x);
    else if (p == 2)
      fill(l, r, x);
    else if (p == 3)
      printf("%I64d\n", kth(l, r, k));
    else if (p == 4)
      printf("%I64d\n", sum(l, r, x, y));
  }
  return 0;
}