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 + 5;
int n, m, vmx;
long long seed;
struct node {
  int l, r;
  mutable long long val;
  bool operator<(const node &x) const { return l < x.l; }
};
set<node> tr;
set<node>::iterator split(int pos) {
  set<node>::iterator iter = tr.lower_bound((node){pos});
  if (iter != tr.end() && iter->l == pos) return iter;
  iter--;
  int L = iter->l, R = iter->r;
  long long Val = iter->val;
  tr.erase(iter);
  tr.insert((node){L, pos - 1, Val});
  return tr.insert((node){pos, R, Val}).first;
}
void assign(int l, int r, long long val) {
  set<node>::iterator R = split(r + 1), L = split(l);
  tr.erase(L, R);
  tr.insert((node){l, r, val});
  return;
}
void add(int l, int r, long long val) {
  set<node>::iterator R = split(r + 1), L = split(l);
  for (; L != R; ++L) L->val = L->val + val;
  return;
}
struct Lsh {
  long long val;
  int sum;
} ls[N];
inline bool Lscmp(Lsh a, Lsh b) { return a.val < b.val; }
inline long long rnk(int l, int r, int x) {
  set<node>::iterator R = split(r + 1), L = split(l);
  int tot = 0;
  for (; L != R; ++L) ls[++tot].val = L->val, ls[tot].sum = (L->r - L->l + 1);
  sort(ls + 1, ls + tot + 1, Lscmp);
  for (int i = 1; i <= tot; i++) {
    x -= ls[i].sum;
    if (x <= 0) return ls[i].val;
  }
}
inline long long ksm(long long a, long long b, long long mod) {
  a %= mod;
  long long ans = 1;
  while (b) {
    if (b & 1) ans = ans * a % mod;
    a = a * a % mod;
    b >>= 1;
  }
  return ans;
}
inline long long calc(int l, int r, int x, int y) {
  set<node>::iterator R = split(r + 1), L = split(l);
  long long ans = 0;
  for (; L != R; ++L)
    ans = (ans + ksm(L->val, x, y) * (L->r - L->l + 1) % y) % y;
  return ans;
}
inline int rnd() {
  int ret = seed;
  seed = (seed * 7 + 13) % 1000000007;
  return ret;
}
int op, l, r, x, y;
int tot;
int main() {
  scanf("%d%d%lld%d", &n, &m, &seed, &vmx);
  for (int i = 1; i <= n; i++) tr.insert((node){i, i, rnd() % vmx + 1});
  tr.insert((node){n + 1, n + 1, 0});
  for (int i = 1; i <= m; i++) {
    op = rnd() % 4 + 1;
    l = rnd() % n + 1;
    r = rnd() % n + 1;
    if (l > r) swap(l, r);
    if (op == 3)
      x = rnd() % (r - l + 1) + 1;
    else
      x = rnd() % vmx + 1;
    if (op == 4) y = rnd() % vmx + 1;
    if (op == 1)
      add(l, r, x);
    else if (op == 2)
      assign(l, r, x);
    else if (op == 3)
      printf("%lld\n", rnk(l, r, x));
    else
      printf("%lld\n", calc(l, r, x, y));
  }
  return 0;
}