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;
class odt {
 public:
  mutable unsigned long long v;
  unsigned long long l, r;
  odt(unsigned long long L, unsigned long long R = -1, unsigned long long V = 0)
      : l(L), r(R), v(V) {}
  bool operator<(const odt& o) const { return l < o.l; }
};
class kth {
 public:
  pair<unsigned long long, unsigned long long> pa1;
  kth(pair<unsigned long long, unsigned long long> pa2) : pa1(pa2) {}
  bool operator<(const kth& _) const { return pa1.first > _.pa1.first; }
};
set<odt> ss;
priority_queue<kth> pri;
unsigned long long seed;
unsigned long long a[100005], y;
namespace BASICMATH {
void inc(unsigned long long& x, unsigned long long y1) { x = (x + y1) % y; }
}  // namespace BASICMATH
using namespace BASICMATH;
unsigned long long read() {
  unsigned long long A;
  bool K;
  char C;
  C = A = K = 0;
  while (C < '0' || C > '9') K |= C == '-', C = getchar();
  while (C > '/' && C < ':') A = (A << 3) + (A << 1) + (C ^ 48), C = getchar();
  return (K ? -A : A);
}
inline set<odt>::iterator split(unsigned long long pos) {
  set<odt>::iterator it = ss.lower_bound(odt(pos));
  if (ss.end() != it && it->l == pos) return it;
  unsigned long long v = (--it)->v;
  unsigned long long l = it->l, r = it->r;
  ss.erase(it);
  ss.insert(odt(l, pos - 1, v));
  return ss.insert(odt(pos, r, v)).first;
}
void assign(unsigned long long l, unsigned long long r,
            unsigned long long val) {
  set<odt>::iterator itr = split(r + 1), itl = split(l);
  ss.erase(itl, itr);
  ss.insert(odt(l, r, val));
}
void add(unsigned long long l, unsigned long long r, unsigned long long val) {
  set<odt>::iterator itr = split(r + 1), itl = split(l);
  for (; itr != itl; ++itl) itl->v += val;
}
unsigned long long query(unsigned long long l, unsigned long long r,
                         long long val) {
  set<odt>::iterator itr = split(r + 1), itl = split(l);
  pair<unsigned long long, unsigned long long> pa;
  while (!pri.empty()) pri.pop();
  for (; itl != itr; ++itl) pri.push(make_pair(itl->v, itl->r - itl->l + 1));
  while (val > 0) {
    pa = pri.top().pa1;
    val -= pa.second;
    pri.pop();
  }
  return pa.first;
}
unsigned long long ksm(unsigned long long x, unsigned long long j) {
  if (!j) return 1;
  x %= y;
  return ksm(x * x % y, j >> 1) * (j & 1 ? x : 1) % y;
}
unsigned long long cifa(unsigned long long l, unsigned long long r,
                        unsigned long long val) {
  set<odt>::iterator itr = split(r + 1), itl = split(l);
  unsigned long long sum = 0;
  for (; itl != itr; ++itl) inc(sum, (itl->r - itl->l + 1) * ksm(itl->v, val));
  return sum;
}
unsigned long long rnd() {
  unsigned long long ret;
  ret = seed;
  seed = (seed * 7 + 13) % 1000000007;
  return ret;
}
signed main() {
  unsigned long long n, m, max, i, op, l, r, vmax, x, t;
  n = read(), m = read(), seed = read(), vmax = read();
  for (i = 1; i <= n; i++)
    a[i] = (rnd() % vmax) + 1, ss.insert(odt(i, i, a[i]));
  ss.insert(odt(n + 1));
  for (i = 1; i <= m; i++) {
    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() % vmax) + 1;
    if (op == 4) y = (rnd() % vmax) + 1;
    if (op == 1)
      add(l, r, x);
    else if (op == 2)
      assign(l, r, x);
    else
      printf("%llu\n", (op == 3 ? query(l, r, x) : cifa(l, r, x)));
  }
  return 0;
}