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;
inline long long read() {
  register long long ret = 0, f = 0;
  char ch = getchar();
  while (!isdigit(ch)) {
    if (ch == '-') f = 1;
    ch = getchar();
  }
  while (isdigit(ch)) {
    ret = ret * 10 + ch - 48;
    ch = getchar();
  }
  return f ? -ret : ret;
}
struct odt {
  long long l, r;
  mutable long long val;
  odt(long long x, long long y, long long z) {
    l = x;
    r = y;
    val = z;
  }
  bool operator<(const odt& x) const { return l < x.l; }
};
set<odt> t;
long long n, m, seed, vm;
long long a[100005], opt, l, r, x, y;
inline long long rd() {
  long long ret = seed;
  seed = (seed * 7 + 13) % 1000000007;
  return ret;
}
inline long long ksm(long long base, long long p, long long m) {
  long long res = 1;
  base %= m;
  while (p) {
    if (p & 1) {
      res *= base;
      res %= m;
    }
    base *= base;
    base %= m;
    p >>= 1;
  }
  return res;
}
inline set<odt>::iterator split(long long x) {
  set<odt>::iterator it = t.lower_bound((odt){x, 0, 0});
  if (it != t.end() && it->l == x) return it;
  --it;
  int sl = it->l, sr = it->r;
  long long sv = it->val;
  t.erase(it);
  t.insert((odt){sl, x - 1, sv});
  return t.insert((odt){x, sr, sv}).first;
}
inline void assi(long long l, long long r, long long v) {
  set<odt>::iterator it1 = split(l), it2 = split(r + 1);
  t.erase(it1, it2);
  t.insert((odt){l, r, v});
}
inline void add(long long l, long long r, long long v) {
  for (set<odt>::iterator it1 = split(l), it2 = split(r + 1); it1 != it2; ++it1)
    it1->val += v;
}
inline long long rk(long long l, long long r, long long v) {
  vector<pair<long long, long long> > tmp;
  tmp.clear();
  for (set<odt>::iterator it1 = split(l), it2 = split(r + 1); it1 != it2;
       ++it1) {
    tmp.push_back((pair<long long, long long>){it1->val, it1->r - it1->l + 1});
  }
  sort(tmp.begin(), tmp.end());
  for (register int i = 0; i < tmp.size(); ++i) {
    v -= tmp[i].second;
    if (v <= 0) return tmp[i].first;
  }
  return -0x7f7f7f7f7f7f7f7f;
}
inline long long spow(long long l, long long r, long long x, long long y) {
  long long res = 0;
  for (set<odt>::iterator it1 = split(l), it2 = split(r + 1); it1 != it2; ++it1)
    res += (it1->r - it1->l + 1) * ksm(it1->val, x, y) % y, res %= y;
  return res;
}
int main() {
  n = read();
  m = read();
  seed = read();
  vm = read();
  for (register int i = 1; i <= n; ++i) {
    a[i] = (rd() % vm) + 1;
    t.insert((odt){i, i, a[i]});
  }
  for (register int i = 1; i <= m; ++i) {
    opt = (rd() % 4) + 1;
    l = (rd() % n) + 1;
    r = (rd() % n) + 1;
    if (l > r) swap(l, r);
    if (opt == 3)
      x = (rd() % (r - l + 1)) + 1;
    else
      x = (rd() % vm) + 1;
    if (opt == 4) y = (rd() % vm) + 1;
    if (opt == 1)
      add(l, r, x);
    else if (opt == 2)
      assi(l, r, x);
    else if (opt == 3)
      printf("%lld\n", rk(l, r, x));
    else
      printf("%lld\n", spow(l, r, x, y));
  }
  return 0;
}