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 N = 1e5 + 100;
const int MOD = 1e9 + 7;
int n, m;
long long seed, MAXV, in[N];
inline long long rnd() {
  long long res = seed;
  seed = (seed * 7 + 13) % MOD;
  return res;
}
struct Node {
  int l, r;
  mutable long long val;
  Node(int L, int R = -1, long long V = 0) : l(L), r(R), val(V) {}
  bool operator<(const Node &x) const { return l < x.l; }
};
set<Node> S;
inline long long pow(long long x, int p, int mod) {
  long long res = 1;
  while (p) {
    if (p & 1) res = (res * x) % mod;
    x = x * x % mod;
    p >>= 1;
  }
  return res;
}
set<Node>::iterator split(int pos) {
  set<Node>::iterator x = S.lower_bound(Node(pos));
  if (x != S.end() && x->l == pos) return x;
  --x;
  int L = x->l, R = x->r;
  long long V = x->val;
  S.erase(x);
  S.insert(Node(L, pos - 1, V));
  return S.insert(Node(pos, R, V)).first;
}
inline void assign_val(int l, int r, long long val) {
  set<Node>::iterator R = split(r + 1), L = split(l);
  S.erase(L, R);
  S.insert((Node){l, r, val});
}
inline void add(int l, int r, long long val) {
  set<Node>::iterator R = split(r + 1), L = split(l);
  for (set<Node>::iterator i = L; i != R; ++i) i->val += val;
}
vector<pair<long long, int> > Q;
inline long long ranks(int l, int r, int k) {
  set<Node>::iterator R = split(r + 1), L = split(l);
  Q.clear();
  for (set<Node>::iterator i = L; i != R; ++i)
    Q.push_back(pair<long long, int>(i->val, i->r - i->l + 1));
  sort(Q.begin(), Q.end());
  for (vector<pair<long long, int> >::iterator i = Q.begin(); i != Q.end();
       ++i) {
    k -= i->second;
    if (k <= 0) return i->first;
  }
  return -1;
}
inline long long sum(int l, int r, int x, int y) {
  set<Node>::iterator R = split(r + 1), L = split(l);
  long long res = 0;
  for (set<Node>::iterator i = L; i != R; ++i)
    res = (res + (long long)(i->r - i->l + 1) * pow(i->val % y, x, y) % y) % y;
  return res;
}
int main() {
  scanf("%d%d%lld%lld", &n, &m, &seed, &MAXV);
  for (int i = 1; i <= n; ++i) {
    in[i] = rnd() % MAXV + 1;
    S.insert((Node){i, i, in[i]});
  }
  S.insert((Node){n + 1, n + 1, 0});
  for (int i = 1; i <= m; ++i) {
    int opt = rnd() % 4 + 1;
    int l = rnd() % n + 1;
    int r = rnd() % n + 1;
    if (l > r) swap(l, r);
    int x, y;
    if (opt == 3)
      x = rnd() % (r - l + 1) + 1;
    else
      x = rnd() % MAXV + 1;
    if (opt == 4) y = rnd() % MAXV + 1;
    if (opt == 1)
      add(l, r, x);
    else if (opt == 2)
      assign_val(l, r, x);
    else if (opt == 3)
      printf("%lld\n", ranks(l, r, x));
    else
      printf("%lld\n", sum(l, r, x, y));
  }
}