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 nod {
  int l, r;
  mutable long long v;
  inline bool operator<(const nod b) const { return l < b.l; }
  nod(int a, int b = -1, long long c = -1) { l = a, r = b, v = c; }
};
int n;
set<nod> li;
set<nod>::iterator split(int p) {
  set<nod>::iterator a = li.lower_bound(nod(p, -1));
  if (a != li.end() && a->l == p) return a;
  --a;
  int l = a->l, r = a->r;
  long long v = a->v;
  li.erase(a);
  li.insert(nod(l, p - 1, v));
  return li.insert(nod(p, r, v)).first;
}
void add(int l, int r, long long x) {
  set<nod>::iterator b = split(r + 1), a = split(l);
  for (set<nod>::iterator i = a; i != b; ++i) {
    (i->v) += x;
  }
}
inline void assign(int l, int r, long long x) {
  set<nod>::iterator b = split(r + 1), a = split(l);
  li.erase(a, b);
  li.insert(nod(l, r, x));
}
long long g_rank(int l, int r, int x) {
  x--;
  vector<pair<long long, int> > v;
  set<nod>::iterator b = split(r + 1), a = split(l);
  for (; a != b; a++) {
    v.push_back(make_pair(a->v, (a->r) - (a->l) + 1));
  }
  sort(v.begin(), v.end());
  vector<pair<long long, int> >::iterator iter = v.begin();
  for (; x >= iter->second; x -= iter->second, iter++)
    ;
  return iter->first;
}
long long gsc(long long a, long long b, long long mod) {
  long long ans = 0;
  a %= mod, b %= mod;
  while (b) {
    if (b & 1) {
      ans = (ans + a) % mod;
    }
    a = (a + a) % mod;
    b >>= 1;
  }
  return ans;
}
long long ksm(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 get_ans(int l, int r, long long x, long long y) {
  set<nod>::iterator b = split(r + 1), a = split(l);
  long long ans = 0;
  for (; a != b; a++) {
    ans += gsc(ksm(a->v, x, y), (a->r) - (a->l) + 1, y);
    ans %= y;
  }
  return ans;
}
void print(int p) {
  set<nod>::iterator a = split(p);
  cout << a->v << endl;
  return;
}
int m;
long long seed, vmax;
inline long long rnd() {
  long long ret = seed;
  seed = (seed * 7 + 13) % 1000000007;
  return ret;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> m >> seed >> vmax;
  for (int i = 1; i <= n; i++) {
    li.insert(nod(i, i, (rnd() % vmax) + 1));
  }
  for (int i = 0; i < m; i++) {
    int op = (rnd() % 4) + 1;
    int a = (rnd() % n) + 1;
    int b = (rnd() % n) + 1;
    if (a > b) swap(a, b);
    if (op == 1) {
      long long c = (rnd() % vmax) + 1;
      add(a, b, c);
    } else if (op == 2) {
      long long c = (rnd() % vmax) + 1;
      assign(a, b, c);
    } else if (op == 3) {
      int c = (rnd() % (b - a + 1)) + 1;
      cout << g_rank(a, b, c) << "\n";
    } else if (op == 4) {
      long long c = (rnd() % vmax) + 1;
      long long d = (rnd() % vmax) + 1;
      cout << get_ans(a, b, c, d) << "\n";
    }
  }
  return 0;
}