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;
const int MOD7 = 1e9 + 7;
const int MOD9 = 1e9 + 9;
const int MAXN = 1e5 + 10;
inline long long qpow(long long a, long long b, long long mod) {
  long long res = 1;
  long long ans = a % mod;
  while (b) {
    if (b & 1) res = res * ans % mod;
    ans = ans * ans % mod;
    b >>= 1;
  }
  return res;
}
struct node {
  int l, r;
  mutable long long v;
  node() {}
  node(int L, int R = -1, long long V = 0) : l(L), r(R), v(V) {}
  bool operator<(const node& o) const { return l < o.l; }
};
set<node> s;
inline set<node>::iterator split(int pos) {
  set<node>::iterator it = s.lower_bound(node(pos));
  if (it != s.end() && it->l == pos) return it;
  --it;
  int L = it->l, R = it->r;
  long long V = it->v;
  s.erase(it);
  s.insert(node(L, pos - 1, V));
  return s.insert(node(pos, R, V)).first;
}
inline void add(int l, int r, long long val = 1) {
  set<node>::iterator itl = split(l), itr = split(r + 1);
  for (; itl != itr; ++itl) itl->v += val;
}
inline void assignval(int l, int r, long long val = 0) {
  set<node>::iterator itl = split(l), itr = split(r + 1);
  s.erase(itl, itr);
  s.insert(node(l, r, val));
}
inline long long rankk(int l, int r, int k) {
  vector<pair<long long, int> > vp;
  set<node>::iterator itl = split(l), itr = split(r + 1);
  vp.clear();
  for (; itl != itr; ++itl)
    vp.push_back(pair<long long, int>(itl->v, itl->r - itl->l + 1));
  sort(vp.begin(), vp.end());
  for (vector<pair<long long, int> >::iterator it = vp.begin(); it != vp.end();
       ++it) {
    k -= it->second;
    if (k <= 0) return it->first;
  }
  return -1LL;
}
inline long long sum(int l, int r, int ex, int mod) {
  set<node>::iterator itl = split(l), itr = split(r + 1);
  long long res = 0;
  for (; itl != itr; ++itl)
    res = (res + (long long)(itl->r - itl->l + 1) *
                     qpow(itl->v, (long long)(ex), (long long)(mod))) %
          mod;
  return res;
}
int n, m;
long long seed, vmax;
long long rnd() {
  long long ret = seed;
  seed = (seed * 7 + 13) % MOD7;
  return ret;
}
long long a[MAXN];
int main() {
  cin >> n >> m >> seed >> vmax;
  for (int i = 1; i <= n; i++) {
    a[i] = (rnd() % vmax) + 1;
    s.insert(node(i, i, a[i]));
  }
  s.insert(node(n + 1, n + 1, 0));
  int lines = 0;
  for (int i = 1; i <= m; i++) {
    int op = int(rnd() % 4) + 1;
    int l = int(rnd() % n) + 1;
    int r = int(rnd() % n) + 1;
    if (l > r) swap(l, r);
    int x, y;
    if (op == 3)
      x = int(rnd() % (r - l + 1)) + 1;
    else
      x = int(rnd() % vmax) + 1;
    if (op == 4) y = int(rnd() % vmax) + 1;
    if (op == 1)
      add(l, r, (long long)(x));
    else if (op == 2)
      assignval(l, r, (long long)(x));
    else if (op == 3)
      printf("%lld\n", rankk(l, r, x));
    else
      printf("%lld\n", sum(l, r, x, y));
  }
  return 0;
}