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;
const int NIL = -1;
template <class T>
inline T mx(T a, T b) {
  return a > b ? a : b;
}
template <class T>
inline T mi(T a, T b) {
  return a < b ? a : b;
}
template <class T>
inline void sw(T &a, T &b) {
  T t = a;
  a = b;
  b = t;
}
template <class T>
inline T mabs(T x) {
  return x < 0 ? -x : x;
}
inline char gc() {
  char ret;
  while ((ret = getchar()) == ' ' || ret == '\n' || ret == '\t')
    ;
  return ret;
}
template <class T>
inline T sq(T x) {
  return x * x;
}
const int LIM = 1e5 + 10;
long long n, m, seed, vmax;
map<long long, long long> ct;
inline map<long long, long long>::iterator split(long long p) {
  map<long long, long long>::iterator ret = --ct.upper_bound(p);
  if (ret->first != p) ct[p] = ret->second, ret++;
  return ret;
}
inline void add(long long l, long long r, long long diff) {
  map<long long, long long>::iterator s = split(l), e = split(r + 1);
  while (s != e) (s++)->second += diff;
}
inline void assign(long long l, long long r, long long x) {
  map<long long, long long>::iterator s = split(l), e = split(r + 1);
  while (s != e) ct.erase(s++);
  ct[l] = x;
}
vector<pair<long long, long long> > buff;
inline long long kSmallest(long long l, long long r, long long k) {
  map<long long, long long>::iterator s = split(l), e = split(r + 1);
  buff.clear();
  while (s != e) {
    r = (e--)->first;
    buff.push_back(make_pair(e->second, r - e->first));
  }
  sort(buff.begin(), buff.end());
  for (int i = 0; i < buff.size(); i++)
    if ((k -= buff[i].second) <= 0) return buff[i].first;
  return -1;
}
inline long long fp(long long base, long long p, long long mod) {
  long long ret = 1;
  base %= mod;
  do {
    if (p & 1) ret = ret * base % mod;
    base = base * base % mod;
  } while (p >>= 1);
  return ret;
}
inline long long powSum(long long l, long long r, long long p, long long mod) {
  map<long long, long long>::iterator s = split(l), e = split(r + 1);
  long long ret = 0;
  while (s != e) {
    r = (e--)->first;
    ret = (ret + (r - e->first) * fp(e->second, p, mod)) % mod;
  }
  return ret % mod;
}
inline long long rnd() {
  long long ret = seed;
  seed = (seed * 7 + 13) % 1000000007;
  return ret;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL), cout.tie(NULL);
  cin >> n >> m >> seed >> vmax;
  long long x, y;
  for (int i = 1; i <= n; i++) ct[i] = (rnd() % vmax) + 1;
  for (int i = 1; i <= m; i++) {
    long long op = (rnd() % 4) + 1;
    long long l = (rnd() % n) + 1;
    long long r = (rnd() % n) + 1;
    if (l > r) sw(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:
        assign(l, r, x);
        break;
      case 3:
        cout << kSmallest(l, r, x) << '\n';
        break;
      case 4:
        cout << powSum(l, r, x, y) << '\n';
        break;
    }
  }
  return 0;
}