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 T1, typename T2>
bool Max(T1& a, T2 b) {
  return a < b ? a = b, 1 : 0;
}
template <typename T1, typename T2>
bool Min(T1& a, T2 b) {
  return a > b ? a = b, 1 : 0;
}
long long pow(long long v, int p, int m) {
  long long res = 1 % m;
  v %= m;
  while (p) {
    if (p & 1) res = res * v % m;
    v = v * v % m;
    p >>= 1;
  }
  return res;
}
const int N = 100001, inf = 1e9 + 1;
long long seed;
long long rnd() {
  long long ret = seed;
  seed = (seed * 7 + 13) % 1000000007;
  return ret;
}
void solve() {
  int n, m, vm;
  cin >> n >> m >> seed >> vm;
  map<int, long long> mp;
  for (int i = 1; i <= (n); ++i) {
    mp[i] = (rnd() % vm) + 1;
  }
  mp[n + 1] = 0;
  while (m--) {
    int op = (rnd() % 4) + 1;
    int l = (rnd() % n) + 1;
    int r = (rnd() % n) + 1;
    if (l > r) swap(l, r);
    r++;
    int x;
    if (op == 3)
      x = (rnd() % (r - l)) + 1;
    else
      x = (rnd() % vm) + 1;
    auto it1 = mp.lower_bound(l);
    if (it1->first != l) {
      long long v = (--it1)->second;
      it1 = mp.insert(++it1, {l, v});
    }
    auto it2 = mp.lower_bound(r);
    if (it2->first != r) {
      long long v = (--it2)->second;
      it2 = mp.insert(++it2, {r, v});
    }
    if (op == 1) {
      while (it1 != it2) {
        (it1++)->second += x;
      }
    } else if (op == 2) {
      while (it1 != it2) mp.erase(it1++);
      mp.insert(it2, {l, x});
    } else if (op == 3) {
      map<long long, int> cnt;
      while (it1 != it2) {
        long long v = it1->second;
        int pos = it1->first;
        it1++;
        cnt[v] += it1->first - pos;
      }
      auto it = cnt.begin();
      while (it->second < x) {
        x -= it->second;
        it++;
      }
      cout << it->first << '\n';
    } else {
      int y = (rnd() % vm) + 1;
      long long res = 0;
      while (it1 != it2) {
        long long v = it1->second;
        int pos = it1->first;
        it1++;
        int cnt = it1->first - pos;
        long long t = pow(v, x, y) * cnt % y;
        res = (res + t) % y;
      }
      cout << res << '\n';
    }
  }
}
void init() {}
int main(void) {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.setf(ios::fixed);
  cout.precision(20);
  init();
  int TC = 1;
  for (int TI = 1; TI <= (TC); ++TI) {
    solve();
  }
  return 0;
}