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;
long long sd;
long long rn() {
  long long ret = sd;
  sd = (sd * 7 + 13) % 1000000007;
  return ret;
}
inline long long rd() {
  long long x = 0, p = 1;
  char a = getchar();
  while ((a < 48 || a > 57) && a != '-') a = getchar();
  if (a == '-') p = -p, a = getchar();
  while (a > 47 && a < 58) {
    x = (x << 1) + (x << 3) + (a & 15);
    a = getchar();
  }
  return x * p;
}
inline void swap(long long &x, long long &y) { x ^= y ^= x ^= y; }
inline long long fpow(long long b, long long p, long long k) {
  long long tmp = b % k, ans = 1;
  while (p) {
    if (p & 1) ans = ans * tmp % k;
    tmp = tmp * tmp % k;
    p >>= 1;
  }
  return ans % k;
}
long long n, q, vm;
struct node {
  long long l, r;
  mutable long long v;
  node(long long tl, long long tr = -1, long long tv = 0) {
    l = tl;
    r = tr;
    v = tv;
  }
  bool operator<(const node &y) const { return l < y.l; }
};
set<node> tree;
inline set<node>::iterator split(long long pos) {
  set<node>::iterator it = tree.lower_bound(node(pos));
  if (it != tree.end() && it->l == pos) return it;
  it--;
  long long tl = it->l;
  long long tr = it->r;
  long long tv = it->v;
  tree.erase(it);
  tree.insert(node(tl, pos - 1, tv));
  return tree.insert(node(pos, tr, tv)).first;
}
inline void assign(long long l, long long r, long long v) {
  set<node>::iterator il = split(l), ir = split(r + 1);
  tree.erase(il, ir);
  tree.insert(node(l, r, v));
}
inline void update(long long l, long long r, long long v) {
  set<node>::iterator il = split(l), ir = split(r + 1);
  for (; il != ir; il++) il->v += v;
}
inline long long kth(long long l, long long r, long long k) {
  vector<pair<long long, long long> > v;
  set<node>::iterator il = split(l), ir = split(r + 1);
  for (; il != ir; il++) v.push_back(make_pair(il->v, il->r - il->l + 1));
  sort(v.begin(), v.end());
  for (vector<pair<long long, long long> >::iterator it = v.begin(),
                                                     ed = v.end();
       it != ed; it++) {
    k -= it->second;
    if (k <= 0) return it->first;
  }
  return -1ll;
}
inline long long squery(long long l, long long r) {
  long long ans = 0;
  set<node>::iterator il = split(l), ir = split(r + 1);
  for (; il != ir; il++) ans += (il->r - il->l + 1) * (il->v);
  return ans;
}
inline long long pquery(long long l, long long r, long long x, long long mod) {
  long long ans = 0;
  set<node>::iterator il = split(l), ir = split(r + 1);
  for (; il != ir; il++)
    ans = (ans + (il->r - il->l + 1) * fpow(il->v, x, mod)) % mod;
  return ans % mod;
}
signed main() {
  n = rd();
  q = rd();
  sd = rd();
  vm = rd();
  for (long long i = 1; i <= n; i++) {
    long long a = rn() % vm + 1;
    tree.insert(node(i, i, a));
  }
  tree.insert(node(n + 1, n + 1, 0));
  while (q--) {
    long long op, l, r, x, y;
    op = rn() % 4 + 1;
    l = rn() % n + 1;
    r = rn() % n + 1;
    if (l > r) swap(l, r);
    if (op == 3)
      x = rn() % (r - l + 1) + 1;
    else
      x = rn() % vm + 1;
    if (op == 4) y = rn() % vm + 1;
    if (op == 1) update(l, r, x);
    if (op == 2) assign(l, r, x);
    if (op == 3) printf("%lld\n", kth(l, r, x));
    if (op == 4) printf("%lld\n", pquery(l, r, x, y));
  }
  return 0;
}