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 _Tp>
inline void IN(_Tp& dig) {
  char c;
  bool flag = 0;
  dig = 0;
  while (c = getchar(), !isdigit(c))
    if (c == '-') flag = 1;
  while (isdigit(c)) dig = dig * 10 + c - '0', c = getchar();
  if (flag) dig = -dig;
}
long long qpow(long long a, long long b, long long mod) {
  long long res = 1;
  a %= mod;
  while (b) {
    if (b & 1) res = res * a % mod;
    a = a * a % mod, b >>= 1;
  }
  return res % mod;
}
struct N {
  int l, r;
  mutable long long d;
  N(int a, int b, long long c) { l = a, r = b, d = c; }
};
list<N> t;
void Divide(int l, int r, list<N>::iterator& L, list<N>::iterator& R) {
  L = t.begin();
  while (L->r < l) L++;
  if (L->l < l) t.insert(L, N(L->l, l - 1, L->d)), L->l = l;
  R = L;
  while (R->r < r) R++;
  if (R->r > r) {
    list<N>::iterator tmp = R;
    tmp++;
    t.insert(tmp, N(r + 1, R->r, R->d)), R->r = r;
  }
  R++;
}
void Add(int l, int r, long long x) {
  list<N>::iterator L, R;
  Divide(l, r, L, R);
  while (L != R) L->d += x, L++;
}
void Modify(int l, int r, long long x) {
  list<N>::iterator L, R;
  Divide(l, r, L, R);
  L->r = r, L->d = x, L++, t.erase(L, R);
}
long long Kth(int l, int r, int x) {
  static vector<pair<long long, int> > srt;
  srt.clear();
  list<N>::iterator L, R;
  Divide(l, r, L, R);
  while (L != R)
    srt.push_back(pair<long long, int>(L->d, L->r - L->l + 1)), L++;
  sort(srt.begin(), srt.end());
  long long res = -1;
  int i = 0;
  while (x > 0) res = srt[i].first, x -= srt[i].second, i++;
  return res;
}
long long PowSum(int l, int r, long long x, long long y) {
  list<N>::iterator L, R;
  Divide(l, r, L, R);
  long long res = 0;
  while (L != R)
    res = (res + qpow(L->d, x, y) * ((L->r - L->l + 1) % y) % y) % y, L++;
  return res;
}
int n, m;
long long seed, vmx;
long long rnd() {
  long long res = seed;
  seed = (seed * 7 + 13) % 1000000007;
  return res;
}
int main(int argc, char const* argv[]) {
  IN(n), IN(m), IN(seed), IN(vmx);
  for (long long i = 1; i <= n; i += 1) t.push_back(N(i, i, rnd() % vmx + 1));
  for (long long i = 1, op, l, r, x; i <= m; i += 1) {
    op = rnd() % 4 + 1, l = rnd() % n + 1, r = rnd() % n + 1;
    if (l > r) swap(l, r);
    if (op == 3)
      x = rnd() % (r - l + 1) + 1;
    else
      x = rnd() % vmx + 1;
    if (op == 1)
      Add(l, r, x);
    else if (op == 2)
      Modify(l, r, x);
    else if (op == 3)
      printf("%lld\n", Kth(l, r, x));
    else if (op == 4)
      printf("%lld\n", PowSum(l, r, x, rnd() % vmx + 1));
  }
  return 0;
}