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;
struct Node {
  int ll, rr;
  long long val;
  bool operator<(const Node& a) const { return ll < a.ll; };
  Node(int l, int r = 0x7fffffff, long long v = -1) : ll(l), rr(r), val(v) {}
};
set<Node> st;
set<Node>::iterator split(int p) {
  set<Node>::iterator it = st.lower_bound(Node(p));
  if (it != st.end() && it->ll == p) return it;
  it--;
  int l = it->ll, r = it->rr;
  long long v = it->val;
  st.erase(it);
  st.insert(Node(l, p - 1, v));
  return st.insert(Node(p, r, v)).first;
}
void add(int l, int r, int a) {
  set<Node>::iterator rt = split(r + 1), lt = split(l);
  vector<Node> nn;
  for (set<Node>::iterator i = lt; i != rt; i++) {
    nn.push_back(Node(i->ll, i->rr, i->val + a));
  }
  st.erase(lt, rt);
  for_each(nn.begin(), nn.end(), [](Node i) { st.insert(i); });
}
void assign(int l, int r, int a) {
  set<Node>::iterator rt = split(r + 1), lt = split(l);
  st.erase(lt, rt);
  st.insert(Node(l, r, a));
}
long long kth(int l, int r, int k) {
  set<Node>::iterator rt = split(r + 1), lt = split(l);
  vector<pair<long long, int>> a;
  while (lt != rt) {
    a.push_back(make_pair(lt->val, lt->rr - lt->ll + 1));
    lt++;
  }
  sort(a.begin(), a.end());
  auto i = a.begin();
  while (k > i->second) {
    k -= i->second;
    i++;
  }
  return i->first;
}
long long qp(long long x, int k, int m) {
  long long ans = 1;
  long long base = x % m;
  while (k) {
    if (k & 1) {
      ans *= base;
      assert(ans >= 0);
      ans %= m;
    }
    base *= base;
    base %= m;
    k >>= 1;
  }
  return ans;
}
int p(int l, int r, int x, int y) {
  set<Node>::iterator rt = split(r + 1), lt = split(l);
  long long ans = 0;
  while (lt != rt) {
    ans += ((qp(lt->val, x, y) % y) * ((lt->rr - lt->ll + 1)) % y) % y;
    ans %= y;
    lt++;
  }
  return ans;
}
int op, l, r, x, y;
int n, m, vmax;
long long seed;
int rnd() {
  int ret = seed;
  seed = (seed * 7 + 13) % 1000000007;
  return ret;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  cin >> n >> m >> seed >> vmax;
  st.insert(Node(0, n, 0));
  for (int i = 0; i < n; i++) {
    assign(i, i, rnd() % vmax + 1);
  }
  for (int i = 0; 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;
    switch (op) {
      case 1:
        add(l - 1, r - 1, x);
        break;
      case 2:
        assign(l - 1, r - 1, x);
        break;
      case 3:
        cout << kth(l - 1, r - 1, x) << endl;
        break;
      case 4:
        cout << p(l - 1, r - 1, x, y) << endl;
        break;
    }
  }
  return 0;
}