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 int read(register int ans = 0, register int sgn = ' ',
                register int ch = getchar()) {
  for (; ch < '0' || ch > '9'; sgn = ch, ch = getchar())
    ;
  for (; ch >= '0' && ch <= '9'; (ans *= 10) += ch - '0', ch = getchar())
    ;
  return sgn - '-' ? ans : -ans;
}
char _Ob[1000000 + 5], *_O = _Ob, _Os[25], *_Ot;
template <typename T>
inline void Out(T x) {
  if (!x) {
    (*_O++ = '0');
    return;
  }
  if (x < 0) (*_O++ = '-'), x = -x;
  for (_Ot = _Os; x; *_Ot++ = x % 10 + '0', x /= 10)
    ;
  for (; _Ot != _Os; (*_O++ = *--_Ot))
    ;
  if (_O - _Ob >= 1000000 - 50) fwrite(_Ob, 1, _O - _Ob, stdout), _O = _Ob;
}
struct D {
  long long v;
  int n;
  D() {}
  D(long long _v, int _n) { v = _v, n = _n; }
} d[100000 + 5];
inline bool operator<(D a, D b) { return a.v < b.v; }
int n, m, op, ql, qr, x, y, top;
long long seed, vmax, ret, ans;
map<int, long long> M;
map<int, long long>::iterator lj, rj, tj;
inline int rd() {
  ret = seed, seed = (7 * seed + 13) % 1000000007;
  return ret;
}
inline void GEN() {
  op = (rd() & 3) + 1, ql = rd() % n + 1, qr = rd() % n + 1,
  ql > qr ? ql ^= qr ^= ql ^= qr : 0,
  x = rd() % (op - 3 ? vmax : qr - ql + 1) + 1,
  op == 4 ? y = rd() % vmax + 1 : 0;
}
inline void GetSeg() {
  rj = M.lower_bound(qr), M.count(qr) ? rj++, 0 : M[qr] = rj->second,
  lj = M.lower_bound(ql - 1), M.count(ql - 1) ? lj++,
  0 : M[ql - 1] = lj->second;
}
inline long long fastpow(long long s, int n, int P) {
  long long a = 1;
  for (; n; n& 1 ? a = a * s % P : 0, s = s * s % P, n >>= 1)
    ;
  return a;
}
int main() {
  n = read(), m = read(), seed = read(), vmax = read(), M[0] = 0;
  for (register int i = 1; i <= n; M[i] = rd() % vmax + 1, i++)
    ;
  M[n + 1] = 0;
  for (register int i; m--;) switch (GEN(), GetSeg(), op) {
      case 1:
        for (; lj != rj; lj->second += x, ++lj)
          ;
        break;
      case 2:
        for (; lj != rj; M.erase(lj++))
          ;
        M[qr] = x;
        break;
      case 3:
        for (tj = lj, --tj; lj != rj;
             d[++top] = D(lj->second, lj->first - tj->first), tj++, lj++)
          ;
        sort(d + 1, d + top + 1);
        for (i = 1; x > d[i].n; x -= d[i].n, i++)
          ;
        Out(d[i].v), (*_O++ = '\n'), top = 0;
        break;
      case 4:
        for (tj = lj, --tj; lj != rj;
             ans = (ans +
                    fastpow(lj->second % y, x, y) * (lj->first - tj->first)) %
                   y,
            tj++, lj++)
          ;
        Out(ans), (*_O++ = '\n');
        ans = 0;
        break;
    }
  fwrite(_Ob, 1, _O - _Ob, stdout), _O = _Ob;
  return 0;
}