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>
const int Mod = 1E9 + 7;
int seed;
inline int Read() {
  int tmp = seed;
  seed = (seed * 7LL + 13) % Mod;
  return tmp;
}
int Pow(int b, int p, int Mod) {
  int res = 1;
  for (b %= Mod; p; p >>= 1, b = 1LL * b * b % Mod)
    if (p & 1) res = 1LL * res * b % Mod;
  return res;
}
class ODT {
 private:
  struct Node {
    int l;
    int r;
    mutable long long v;
    bool operator<(const Node& rhs) const { return l < rhs.l; }
    Node(int l, int r = 0, long long v = 0) : l(l), r(r), v(v) {}
  };
  std::set<Node> s;
  std::set<Node>::iterator Split(int p) {
    std::set<Node>::iterator it = s.lower_bound(Node(p));
    if (it != s.end() && it->l == p) return it;
    --it;
    int l = it->l, r = it->r;
    long long v = it->v;
    s.erase(it);
    s.insert(Node(l, p - 1, v));
    return s.insert(Node(p, r, v)).first;
  }

 public:
  void Cover(int l, int r, long long v) {
    std::set<Node>::iterator R = Split(r + 1), L = Split(l);
    s.erase(L, R);
    s.insert(Node(l, r, v));
  }
  void Add(int l, int r, int v) {
    std::set<Node>::iterator R = Split(r + 1), L = Split(l);
    for (; L != R; ++L) L->v += v;
  }
  long long Kth(int l, int r, int k) {
    std::set<Node>::iterator R = Split(r + 1), L = Split(l);
    std::map<long long, int> f;
    for (; L != R; ++L) f[L->v] += L->r - L->l + 1;
    for (auto it = f.begin(); it != f.end(); ++it)
      if (k <= it->second)
        return it->first;
      else
        k -= it->second;
  }
  int Pow(int l, int r, int x, int y) {
    int ans = 0;
    std::set<Node>::iterator R = Split(r + 1), L = Split(l);
    for (; L != R; ++L)
      ans = (ans + 1LL * (L->r - L->l + 1) * ::Pow(L->v % y, x, y) % y) % y;
    return ans;
  }
  void Insert(int l, int r, long long v) { s.insert(Node(l, r, v)); }
} t;
int main() {
  int n, m, vmax;
  scanf("%d%d%d%d", &n, &m, &seed, &vmax);
  for (int i = 1, x; i <= n; i++) x = (Read() % vmax) + 1, t.Insert(i, i, x);
  for (int i = 1, op, l, r, x, y; i <= m; i++) {
    op = Read() % 4 + 1;
    l = Read() % n + 1;
    r = Read() % n + 1;
    if (l > r) std::swap(l, r);
    if (op == 1)
      x = Read() % vmax + 1, t.Add(l, r, x);
    else if (op == 2)
      x = Read() % vmax + 1, t.Cover(l, r, x);
    else if (op == 3)
      x = Read() % (r - l + 1) + 1, printf("%lld\n", t.Kth(l, r, x));
    else
      x = Read() % vmax + 1, y = Read() % vmax + 1,
      printf("%d\n", t.Pow(l, r, x, y));
  }
  return 0;
}