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 long long MOD = 1e9 + 7;
const double oula = 0.57721566490153286060651209;
using namespace std;
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& t) const { return l < t.l; }
};
set<node> se;
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;
    }
    b >>= 1;
    a = (a * a) % mod;
  }
  return ans;
}
set<node>::iterator split(int pos) {
  set<node>::iterator it = se.lower_bound(node(pos));
  if (it != se.end() && it->l == pos) return it;
  it--;
  int L = it->l, R = it->r;
  long long V = it->val;
  se.erase(it);
  se.insert(node(L, pos - 1, V));
  return se.insert(node(pos, R, V)).first;
}
void assign(int l, int r, long long v) {
  set<node>::iterator itr = split(r + 1), itl = split(l);
  se.erase(itl, itr);
  se.insert(node(l, r, v));
}
void add(int l, int r, long long v) {
  set<node>::iterator itr = split(r + 1);
  for (set<node>::iterator itl = split(l); itl != itr; itl++) {
    itl->val += v;
  }
}
long long Rank(int l, int r, int v) {
  set<node>::iterator itr = split(r + 1);
  vector<pair<long long, int> > tmp;
  for (set<node>::iterator itl = split(l); itl != itr; itl++) {
    tmp.push_back({itl->val, itl->r - itl->l + 1});
  }
  sort(tmp.begin(), tmp.end());
  for (int i = 0; i < tmp.size(); i++) {
    v -= tmp[i].second;
    if (v <= 0) {
      return tmp[i].first;
    }
  }
}
long long query(int l, int r, long long x, long long y) {
  set<node>::iterator itr = split(r + 1);
  long long ans = 0;
  for (set<node>::iterator itl = split(l); itl != itr; itl++) {
    ans = (ans + ((ksm(itl->val, x, y) * (itl->r - itl->l + 1)) % y)) % y;
  }
  return ans;
}
long long n, m, seed, vmax;
long long rnd() {
  long long ans = seed;
  seed = (seed * 7 + 13) % MOD;
  return ans;
}
int main() {
  std::ios::sync_with_stdio(false);
  cin >> n >> m >> seed >> vmax;
  long long x, y;
  for (int i = 1; i <= n; i++) {
    x = (rnd() % vmax) + 1;
    se.insert(node(i, i, x));
  }
  int opt, L, R;
  for (int i = 1; i <= m; i++) {
    opt = (rnd() % 4) + 1;
    L = (rnd() % n) + 1;
    R = (rnd() % n) + 1;
    if (L > R) swap(L, R);
    if (opt == 3) {
      x = (rnd() % (R - L + 1)) + 1;
    } else {
      x = (rnd() % vmax) + 1;
    }
    if (opt == 4) {
      y = (rnd() % vmax) + 1;
    }
    if (opt == 1) {
      add(L, R, x);
    } else if (opt == 2) {
      assign(L, R, x);
    } else if (opt == 3) {
      cout << Rank(L, R, x) << endl;
    } else {
      cout << query(L, R, x, y) << endl;
    }
  }
}