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;
int seed, n, m, vmax;
int rnd() {
  int ret = seed;
  seed = (1LL * seed * 7 + 13) % 1000000007;
  return ret;
}
struct prz {
  int l, r;
  long long x;
  prz() {}
  prz(int a, int b, long long c) : l{a}, r{b}, x{c} {}
};
bool operator<(const prz& a, const prz& other) {
  if (a.l != other.l) return a.l < other.l;
  return a.r < other.r;
}
bool cmp(const prz& a, const prz& b) { return a.x < b.x; }
set<prz> a;
void cut(int x) {
  auto it = prev(a.upper_bound(prz(x, n + 3, -1)));
  if (it->l == x) return;
  auto val = *it;
  a.erase(it);
  a.emplace(val.l, x, val.x);
  a.emplace(x, val.r, val.x);
}
vector<prz> buf;
void gather(int l, int r) {
  cut(l);
  cut(r);
  buf.clear();
  auto it = a.lower_bound(prz(l, -1, -1));
  while (it != a.end() and it->r <= r) {
    buf.push_back(*it);
    it++;
  }
}
long long pow(long long a, int b, int mod) {
  a %= mod;
  long long res = 1;
  while (b > 0) {
    if (b & 1) res = (res * a) % mod;
    a = (a * a) % mod;
    b >>= 1;
  }
  return res;
}
int main() {
  cin >> n >> m >> seed >> vmax;
  for (int i = 0; i < n; i++) a.emplace(i + 1, i + 2, rnd() % vmax + 1);
  while (m--) {
    int op = (rnd() % 4) + 1;
    int l = rnd() % n + 1;
    int r = rnd() % n + 1;
    if (l > r) swap(l, r);
    int x;
    int y;
    if (op == 3)
      x = (rnd() % (r - l + 1)) + 1;
    else
      x = rnd() % vmax + 1;
    if (op == 4) y = rnd() % vmax + 1;
    gather(l, r + 1);
    if (op == 1) {
      for (auto e : buf) {
        a.erase(e);
        a.emplace(e.l, e.r, e.x + x);
      }
    } else if (op == 2) {
      for (auto e : buf) a.erase(e);
      a.emplace(l, r + 1, x);
    } else if (op == 3) {
      sort(buf.begin(), buf.end(), cmp);
      x--;
      for (auto e : buf) {
        if (x < e.r - e.l) {
          cout << e.x << "\n";
          break;
        }
        x -= e.r - e.l;
      }
    } else if (op == 4) {
      long long ans = 0;
      for (auto e : buf) {
        ans += 1LL * pow(e.x, x, y) * (e.r - e.l);
        ans %= y;
      }
      ans = (ans + y) % y;
      cout << ans << "\n";
    }
  }
}