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;
template <typename T>
inline void read(T& x) {
  T f = 1;
  x = 0;
  char ch = getchar();
  while (0 == isdigit(ch)) {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (0 != isdigit(ch)) x = (x << 1) + (x << 3) + ch - '0', ch = getchar();
  x *= f;
}
template <typename T>
inline void write(T x) {
  if (x < 0) {
    x = ~(x - 1);
    putchar('-');
  }
  if (x > 9) write(x / 10);
  putchar(x % 10 + '0');
}
const int inf = 0x3f3f3f3f;
const int N = 1e6 + 100;
struct Node {
  int l, r;
  mutable long long val;
  Node(int l, int r, long long val) : l(l), r(r), val(val) {}
  Node(int l) : l(l) {}
  bool operator<(const Node& t) const { return l < t.l; }
};
set<Node> st;
set<Node>::iterator split(int pos) {
  auto it = st.lower_bound(Node(pos));
  if (it != st.end() && it->l == pos) return it;
  it--;
  int l = it->l, r = it->r;
  long long val = it->val;
  st.erase(it);
  st.insert(Node(l, pos - 1, val));
  return st.insert(Node(pos, r, val)).first;
}
void assign(int l, int r, long long val) {
  auto itr = split(r + 1), itl = split(l);
  st.erase(itl, itr);
  st.insert(Node(l, r, val));
}
void add(int l, int r, long long val) {
  auto itr = split(r + 1), itl = split(l);
  for (auto it = itl; it != itr; it++) it->val += val;
}
long long kth(int l, int r, int k) {
  auto itr = split(r + 1), itl = split(l);
  vector<pair<long long, int>> node;
  for (auto it = itl; it != itr; it++)
    node.emplace_back(it->val, it->r - it->l + 1);
  sort(node.begin(), node.end());
  for (auto it : node) {
    k -= it.second;
    if (k <= 0) return it.first;
  }
  return -1;
}
long long q_pow(long long a, long long b, long long mod) {
  long long ans = 1;
  a %= mod;
  while (b) {
    if (b & 1) ans = ans * a % mod;
    a = a * a % mod;
    b >>= 1;
  }
  return ans;
}
long long query(int l, int r, int x, int mod) {
  auto itr = split(r + 1), itl = split(l);
  long long ans = 0;
  for (auto it = itl; it != itr; it++)
    ans = (ans + q_pow(it->val, x, mod) * (it->r - it->l + 1)) % mod;
  return ans;
}
long long seed;
int rnd() {
  int ret = seed;
  seed = (seed * 7 + 13) % 1000000007;
  return ret;
}
int main() {
  int n, m, vmax;
  read(n), read(m), read(seed), read(vmax);
  for (int i = 1; i <= n; i++) {
    int num = rnd() % vmax + 1;
    st.insert(Node(i, i, num));
  }
  st.insert(Node(n + 1, n + 1, 0));
  for (int i = 1; i <= m; i++) {
    int op = rnd() % 4 + 1, l = rnd() % n + 1, r = rnd() % n + 1, x, y;
    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 if (op == 3)
      write(kth(l, r, x)), putchar('\n');
    else if (op == 4)
      write(query(l, r, x, y)), putchar('\n');
  }
  return 0;
}