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;
void read(long long &x) {
  x = 0;
  char c = getchar();
  int f = 1;
  while (!(c >= '0' && c <= '9')) {
    if (c == '-') f = -1;
    c = getchar();
  }
  while ((c >= '0' && c <= '9')) {
    x = (x << 1) + (x << 3) + (c ^ 48);
    c = getchar();
  }
  x *= f;
}
struct node {
  long long l, r;
  mutable long long val;
  node(long long L) : l(L) {}
  node(long long L, long long R, long long VAL) : l(L), r(R), val(VAL) {}
};
bool operator<(node x, node y) { return x.l < y.l; }
set<node> s;
set<node>::iterator split(long long pos) {
  set<node>::iterator it = s.lower_bound(node(pos));
  if (it != s.end() && it->l == pos) return it;
  it--;
  long long l = it->l, r = it->r;
  long long val = it->val;
  s.erase(it);
  s.insert(node(l, pos - 1, val));
  return s.insert(node(pos, r, val)).first;
}
void add(long long l, long long r, long long val) {
  set<node>::iterator it2 = split(r + 1), it1 = split(l);
  for (set<node>::iterator it = it1; it != it2; ++it) it->val += val;
}
long long kth(long long l, long long r, long long k) {
  set<node>::iterator it2 = split(r + 1), it1 = split(l);
  vector<pair<long long, long long>> v;
  v.clear();
  for (set<node>::iterator it = it1; it != it2; it++)
    v.push_back(pair<long long, long long>(it->val, it->r - it->l + 1));
  sort(v.begin(), v.end());
  for (long long i = 0; i < v.size(); ++i) {
    k -= v[i].second;
    if (k <= 0) return v[i].first;
  }
}
long long ksm(long long a, long long b, long long c) {
  long long ans = 1;
  a %= c;
  while (b) {
    if (b & 1) ans = (ans * a) % c;
    a = (a * a) % c;
    b >>= 1;
  }
  return ans;
}
long long query(long long l, long long r, long long x, long long y) {
  set<node>::iterator it2 = split(r + 1), it1 = split(l);
  long long res = 0;
  for (set<node>::iterator it = it1; it != it2; ++it) {
    res = (res + (it->r - it->l + 1) * ksm(it->val, x, y)) % y;
  }
  return res;
}
void assign(long long l, long long r, long long val) {
  set<node>::iterator it2 = split(r + 1), it1 = split(l);
  s.erase(it1, it2);
  s.insert(node(l, r, val));
}
long long n, m, seed, vmax;
long long rnd() {
  long long ret = seed;
  seed = (seed * 7 + 13) % 1000000007;
  return ret;
}
int main() {
  read(n), read(m), read(seed), read(vmax);
  for (long long i = 1; i <= n; ++i) {
    long long tmp = (rnd() % vmax) + 1;
    s.insert(node(i, i, tmp));
  }
  long long op, l, r, x, y;
  for (long long i = 1; i <= m; ++i) {
    op = (rnd() % 4) + 1;
    l = (rnd() % n) + 1;
    r = (rnd() % n) + 1;
    if (l > r) swap(l, r);
    if (op == 1) {
      x = (rnd() % vmax) + 1;
      add(l, r, x);
    } else if (op == 2) {
      x = (rnd() % vmax) + 1;
      assign(l, r, x);
    } else if (op == 3) {
      x = (rnd() % (r - l + 1)) + 1;
      printf("%lld\n", kth(l, r, x));
    } else {
      x = (rnd() % vmax) + 1;
      y = (rnd() % vmax) + 1;
      printf("%lld\n", query(l, r, x, y));
    }
  }
  return 0;
}