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>
long long n, m, seed, vmax;
long long qpow(long long x, long long y, long long mod) {
  long long res = 1;
  x %= mod;
  while (y) {
    if (y & 1) res = res * x % mod;
    y >>= 1;
    x = x * x % mod;
  }
  return res;
}
class ChthollyTree {
 public:
  class Node {
   public:
    long long l, r;
    mutable long long val;
    Node(long long _l = 0, long long _r = 0, long long _v = 0)
        : l(_l), r(_r), val(_v) {}
    bool operator<(const Node &a) const { return l < a.l; }
  };
  std::set<Node> s;
  long long n;
  void init() { s.clear(); }
  void init(long long _n) {
    s.clear();
    s.insert(Node(1, n = _n));
  }
  std::set<Node>::iterator split(long long p) {
    if (p > n) return s.end();
    std::set<Node>::iterator it = s.lower_bound(Node(p));
    if (it != s.end() && it->l == p) return it;
    --it;
    Node tmp = *it;
    s.erase(it);
    s.insert(Node(tmp.l, p - 1, tmp.val));
    return s.insert(Node(p, tmp.r, tmp.val)).first;
  }
  void assign(long long l, long long r, long long v) {
    std::set<Node>::iterator pr = split(r + 1), pl = split(l);
    s.erase(pl, pr);
    s.insert(Node(l, r, v));
  }
  void inc(long long l, long long r, long long v) {
    std::set<Node>::iterator pr = split(r + 1), pl = split(l);
    for (std::set<Node>::iterator i = pl; i != pr; ++i) i->val += v;
  }
  long long powerSum(long long l, long long r, long long x, long long mod) {
    std::set<Node>::iterator pr = split(r + 1), pl = split(l);
    long long res = 0;
    for (std::set<Node>::iterator i = pl; i != pr; ++i)
      res = (res + qpow(i->val, x, mod) * (i->r - i->l + 1)) % mod;
    return res;
  }
  long long kth(long long l, long long r, long long k) {
    static std::map<long long, long long> h;
    h.clear();
    std::set<Node>::iterator pr = split(r + 1), pl = split(l);
    for (std::set<Node>::iterator i = pl; i != pr; ++i)
      h[i->val] += i->r - i->l + 1;
    for (std::map<long long, long long>::iterator i = h.begin(); i != h.end();
         ++i)
      if ((k -= i->second) <= 0) return i->first;
    return -1;
  }
} odt;
long long rnd() {
  long long res = seed;
  seed = (seed * 7 + 13) % 1000000007;
  return res;
}
int main() {
  std::ios::sync_with_stdio(false);
  std::cin >> n >> m >> seed >> vmax;
  odt.init(n);
  for (long long i = 1; i <= n; ++i) odt.assign(i, i, rnd() % vmax + 1);
  while (m--) {
    long long op = rnd() % 4 + 1, l = rnd() % n + 1, r = rnd() % n + 1, x, y;
    if (l > r) std::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)
      odt.inc(l, r, x);
    else if (op == 2)
      odt.assign(l, r, x);
    else if (op == 3)
      std::cout << odt.kth(l, r, x) << std::endl;
    else
      std::cout << odt.powerSum(l, r, x, y) << std::endl;
  }
  return 0;
}