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;
inline int readInt() {
  char c;
  int tmp = 0, x = 1;
  c = getchar();
  while (c > '9' || c < '0') {
    if (c == '-') x = -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9') {
    tmp = tmp * 10 + c - '0';
    c = getchar();
  }
  return tmp * x;
}
const int maxn = (int)1e5 + 5;
int n, m, seed, vmax, a[maxn], ret;
int rnd() {
  ret = seed;
  seed = ((long long)seed * 7 + 13) % 1000000007;
  return ret;
}
struct node {
  int l, r;
  mutable long long val;
  bool operator<(const node& oth) const { return l < oth.l; }
  node(int l = 0, int r = 0, long long val = 0) : l(l), r(r), val(val) {}
};
set<node> intev;
void split(int pos) {
  set<node>::iterator ite = intev.lower_bound(node(pos, -1, -1));
  if (ite == intev.end() || ite->l > pos) {
    --ite;
    int L = ite->l, R = ite->r;
    long long val = ite->val;
    intev.erase(ite);
    intev.insert(node(L, pos - 1, val));
    intev.insert(node(pos, R, val));
  }
}
long long quickpow(long long x, int k, long long mo) {
  long long ret = 1;
  while (k > 0) {
    if (k & 1) ret = (ret * x) % mo;
    x = x * x % mo;
    k >>= 1;
  }
  return ret;
}
vector<pair<long long, int> > par;
int main() {
  n = readInt(), m = readInt(), seed = readInt(), vmax = readInt();
  for (int i = 1; i <= n; i++) {
    a[i] = rnd() % vmax + 1;
    intev.insert(node(i, i, 1ll * a[i]));
  }
  int op, l, r, x, y;
  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() % vmax + 1;
    if (op == 4) y = (rnd() % vmax) + 1;
    split(l);
    if (r <= n - 1) split(r + 1);
    set<node>::iterator iteL = intev.lower_bound(node(l, -1, -1));
    set<node>::iterator iteR = intev.upper_bound(node(r, -1, -1));
    if (op == 1) {
      for (set<node>::iterator ite = iteL; ite != iteR; ++ite) ite->val += x;
    } else if (op == 2) {
      intev.erase(iteL, iteR);
      intev.insert(node(l, r, x));
    } else if (op == 3) {
      par.clear();
      for (set<node>::iterator ite = iteL; ite != iteR; ++ite)
        par.push_back(make_pair(ite->val, ite->r - ite->l + 1));
      sort(par.begin(), par.end());
      for (int i = 0; i < (int)par.size(); i++) {
        x -= par[i].second;
        if (x <= 0) {
          printf("%lld\n", par[i].first);
          break;
        }
      }
      assert(x <= 0);
    } else if (op == 4) {
      long long ans = 0;
      for (set<node>::iterator ite = iteL; ite != iteR; ++ite) {
        long long valu = quickpow(ite->val % (1ll * y), x, y) % y;
        valu = valu * (ite->r - ite->l + 1) % y;
        ans = (ans + valu) % y;
      }
      printf("%lld\n", ans);
    }
  }
  return 0;
}