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 N = 100005;
struct node {
  long long l, r, *v;
  bool operator<(const node& rhs) const { return l < rhs.l; }
};
std::set<node> root;
long long p(long long x, long long y, long long z) {
  long long ret = 1;
  x %= z;
  for (; y; y >>= 1, x = x * x % z)
    if (y & 1) ret = ret * x % z;
  return ret;
}
std::set<node>::iterator split(long long l) {
  std::set<node>::iterator it = root.lower_bound((node){l, 0, 0});
  if (it != root.end() && it->l == l) return it;
  it--;
  long long L = it->l, R = it->r, V = *(it->v);
  root.erase(it);
  root.insert((node){L, l - 1, new long long(V)});
  return root.insert((node){l, R, new long long(V)}).first;
}
void add(long long l, long long r, long long x) {
  split(l);
  std::set<node>::iterator ub = split(r + 1);
  for (std::set<node>::iterator i = root.lower_bound((node){l, 0, 0}); i != ub;
       i++) {
    *(i->v) += x;
  }
}
void set(long long l, long long r, long long x) {
  split(l);
  std::set<node>::iterator ub = split(r + 1);
  root.erase(root.lower_bound((node){l, 0, 0}), ub);
  root.insert((node){l, r, new long long(x)});
}
long long rank(long long l, long long r, long long x) {
  std::vector<std::pair<long long, long long> > vec;
  split(l);
  std::set<node>::iterator ub = split(r + 1);
  for (std::set<node>::iterator i = root.lower_bound((node){l, 0, 0}); i != ub;
       i++) {
    vec.push_back(std::make_pair(*(i->v), i->r - i->l + 1));
  }
  sort(vec.begin(), vec.end());
  for (std::vector<std::pair<long long, long long> >::iterator it = vec.begin();
       it != vec.end(); it++) {
    if ((x -= it->second) <= 0) return it->first;
  }
  return -1;
}
long long psum(long long l, long long r, long long x, long long y) {
  long long ret = 0;
  split(l);
  std::set<node>::iterator ub = split(r + 1);
  for (std::set<node>::iterator i = root.lower_bound((node){l, 0, 0}); i != ub;
       i++) {
    (ret += (i->r - i->l + 1) * p(*(i->v), x, y)) %= y;
  }
  return ret;
}
long long n, m, seed, vmax;
long long a[N];
long long rnd() {
  long long ret = seed;
  seed = (seed * 7 + 13) % 1000000007;
  return ret;
}
signed main() {
  scanf("%I64d%I64d%I64d%I64d", &n, &m, &seed, &vmax);
  for (long long i = 1; i <= n; i++) {
    a[i] = rnd() % vmax + 1;
    root.insert((node){i, i, a + i});
  }
  root.insert((node){n + 1, n + 1, a + n + 1});
  while (m--) {
    long long op = rnd() % 4 + 1;
    long long l = rnd() % n + 1;
    long long r = rnd() % n + 1;
    long long x, y = 0;
    if (l > r) std::swap(l, r);
    if (op == 3)
      x = rnd() % (r - l + 1) + 1;
    else
      x = rnd() % vmax + 1;
    if (op == 4) y = rnd() % vmax + 1;
    switch (op) {
      case 1:
        add(l, r, x);
        break;
      case 2:
        set(l, r, x);
        break;
      case 3:
        printf("%I64d\n", rank(l, r, x));
        break;
      case 4:
        printf("%I64d\n", psum(l, r, x, y));
        break;
    }
  }
}