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;
namespace Maths {
long long fastpow(long long b, long long i, long long m) {
  long long ret = 1ll;
  b %= m;
  for (; i; b = b * b % m, i >>= 1)
    if (i & 1) ret = ret * b % m;
  return ret;
}
}  // namespace Maths
const int N = 1e5 + 5;
namespace ODT {
struct Node {
  int l, r;
  mutable long long val;
  Node(int _l, int _r = -1, long long _v = 0ll) { l = _l, r = _r, val = _v; }
  bool operator<(const Node &t) const { return l < t.l; }
};
set<Node> odt;
void reset(long long *s, long long *t) {
  odt.clear();
  for (long long *p = s; p != t; p++)
    odt.insert(Node(p - s + 1, p - s + 1, *p));
  odt.insert(Node(t - s + 1, t - s + 1, 0));
}
set<Node>::iterator split(int p) {
  set<Node>::iterator it = odt.lower_bound(Node(p));
  if (it == odt.end() || it->l == p)
    return it;
  else
    it--;
  int l = it->l, r = it->r;
  long long val = it->val;
  odt.erase(it), odt.insert(Node(l, p - 1, val));
  return odt.insert(Node(p, r, val)).first;
}
void assign(int l, int r, long long val) {
  set<Node>::iterator ity = split(r + 1), itx = split(l);
  odt.erase(itx, ity), odt.insert(Node(l, r, val));
}
void add(int l, int r, long long val) {
  set<Node>::iterator ity = split(r + 1), itx = split(l);
  for (; itx != ity; itx++) itx->val += val;
}
long long find_kth(int l, int r, int k) {
  set<Node>::iterator ity = split(r + 1), itx = split(l);
  vector<pair<long long, int> > dat;
  for (; itx != ity; itx++)
    dat.push_back(pair<long long, int>(itx->val, itx->r - itx->l + 1));
  sort(dat.begin(), dat.end());
  for (vector<pair<long long, int> >::iterator it = dat.begin();
       it != dat.end(); it++) {
    k -= it->second;
    if (k <= 0) return it->first;
  }
  return -1;
}
long long power_sum(int l, int r, long long x, long long y) {
  set<Node>::iterator ity = split(r + 1), itx = split(l);
  long long ans = 0ll;
  for (; itx != ity; itx++)
    ans =
        (ans +
         (long long)(itx->r - itx->l + 1) * Maths::fastpow(itx->val, x, y) % y +
         y) %
        y;
  return ans;
}
}  // namespace ODT
long long seed, vmax;
long long A[N];
int n, m;
inline long long rnd() {
  long long ret = seed;
  seed = (seed * 7 + 13) % (long long)(1e9 + 7);
  return ret;
}
signed main() {
  cin >> n >> m >> seed >> vmax;
  for (register int i = 1; i <= n; i++) A[i] = rnd() % vmax + 1;
  ODT::reset(A + 1, A + 1 + n);
  while (m--) {
    long long op, l, r, x, y;
    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;
    switch (op) {
      case 1:
        ODT::add(l, r, x);
        break;
      case 2:
        ODT::assign(l, r, x);
        break;
      case 3:
        cout << ODT::find_kth(l, r, x) << endl;
        break;
      case 4:
        cout << ODT::power_sum(l, r, x, y) << endl;
        break;
    }
  }
  return 0;
}