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>
inline void cmin(int &x, int y) {
  if (y < x) x = y;
}
inline void cmax(int &x, int y) {
  if (y > x) x = y;
}
using namespace std;
long long Power(long long Base, long long Exp, long long Mod) {
  Base %= Mod;
  long long res(1);
  do {
    if (Exp & 1) (res *= Base) %= Mod;
    (Base *= Base) %= Mod;
  } while (Exp >>= 1);
  return res;
}
long long seed;
int rnd() {
  int res(seed);
  seed = (seed * 7 + 13) % 1000000007;
  return res;
}
int N, M, V_Max;
map<int, long long> A;
int main() {
  cin >> N >> M >> seed >> V_Max;
  for (int i(1); i <= N; ++i)
    A.insert(A.end(), make_pair(i, rnd() % V_Max + 1));
  A.insert(A.end(), make_pair(N + 1, 0));
  while (M--) {
    int op((rnd() % 4) + 1);
    int l((rnd() % N) + 1);
    int r((rnd() % N) + 1);
    int x, y;
    if (l > r) swap(l, r);
    if (op == 3)
      x = (rnd() % (r - l + 1)) + 1;
    else
      x = (rnd() % V_Max) + 1;
    if (op == 4) y = (rnd() % V_Max) + 1;
    auto it_l(--A.upper_bound(l));
    if (it_l->first != l) A[l] = it_l->second, ++it_l;
    auto it_r(--A.upper_bound(r + 1));
    if (it_r->first != r + 1) A[r + 1] = it_r->second, ++it_r;
    switch (op) {
      static vector<pair<long long, int>> v;
      static long long res;
      case 1:
        while (it_l != it_r) (--it_r)->second += x;
        break;
      case 2:
        while (it_l != it_r) A.erase(it_l++);
        A[l] = x;
        break;
      case 3:
        v.clear();
        for (int ub; it_l != it_r;)
          ub = (it_r--)->first,
          v.push_back(make_pair(it_r->second, ub - it_r->first));
        sort(v.begin(), v.end());
        for (int i(0), cnt(0);; ++i)
          if (x <= (cnt += v[i].second)) {
            printf("%I64d\n", v[i].first);
            break;
          }
        break;
      case 4:
        res = 0;
        for (int ub; it_l != it_r;)
          ub = (it_r--)->first,
          (res += Power(it_r->second, x, y) * (ub - it_r->first)) %= y;
        printf("%I64d\n", res);
        break;
    }
  }
  return 0;
}