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 long long max(long long x, long long y) { return x > y ? x : y; }
inline long long min(long long x, long long y) { return x < y ? x : y; }
inline long long gcd(long long n, long long m) { return m ? gcd(m, n % m) : n; }
inline long long lcm(long long x, long long y) { return x / gcd(x, y) * y; }
inline void swap(long long &x, long long &y) {
  long long t = x;
  x = y;
  y = t;
  return;
}
namespace yspm {
inline long long read() {
  long long res = 0, f = 1;
  char k;
  while (!isdigit(k = getchar()))
    if (k == '-') f = -1;
  while (isdigit(k)) res = res * 10 + k - '0', k = getchar();
  return res * f;
}
const long long N = 1e5 + 10;
vector<pair<long long, long long> > ton;
long long n, Q, seed, vmax, a[N];
struct node {
  long long l, r;
  mutable long long val;
  node() {}
  node(long long L, long long rr = -1, long long vv = 0) {
    l = L;
    r = rr;
    val = vv;
    return;
  }
  bool operator<(const node &a) const { return l < a.l; }
};
set<node> st;
inline set<node>::iterator split(long long pos) {
  set<node>::iterator it = st.lower_bound(node(pos, 0, -1));
  if (it != st.end() && it->l == pos) return it;
  --it;
  long long tl = it->l, tr = it->r, v = it->val;
  st.erase(it);
  st.insert(node(tl, pos - 1, v));
  return st.insert(node(pos, tr, v)).first;
}
inline void assign(long long l, long long r, long long val = 0) {
  set<node>::iterator R = split(r + 1), L = split(l);
  st.erase(L, R);
  st.insert(node(l, r, val));
  return;
}
inline void add(long long l, long long r, long long v) {
  set<node>::iterator R = split(r + 1), L = split(l);
  while (L != R) L->val += v, ++L;
  return;
}
inline long long rank(long long l, long long r, long long k) {
  set<node>::iterator R = split(r + 1), L = split(l);
  ton.clear();
  while (L != R) ton.push_back(make_pair(L->val, L->r - L->l + 1)), ++L;
  sort(ton.begin(), ton.end());
  for (auto p : ton)
    if (k <= p.second)
      return p.first;
    else
      k -= p.second;
  assert(false);
}
long long mod;
inline long long add(long long x, long long y) {
  return x + y >= mod ? x + y - mod : x + y;
}
inline long long del(long long x, long long y) {
  return x + y < 0 ? x - y + mod : x - y;
}
inline long long mul(long long x, long long y) {
  return x * y - x * y / mod * mod;
}
inline long long ksm(long long x, long long y) {
  x %= mod;
  long long res = 1;
  for (; y; y >>= 1, x = mul(x, x))
    if (y & 1) res = mul(res, x);
  return res;
}
inline long long getrand() {
  long long res = seed;
  seed = (seed * 7 + 13) % 1000000007;
  return res;
}
inline long long query(long long l, long long r, long long x) {
  set<node>::iterator R = split(r + 1), L = split(l);
  long long ans = 0;
  while (L != R) ans = add(ans, mul(L->r - L->l + 1, ksm(L->val, x))), ++L;
  return ans;
}
signed main() {
  n = read();
  Q = read();
  seed = read();
  vmax = read();
  for (register long long i = 1; i <= n; ++i)
    a[i] = getrand() % vmax + 1, st.insert(node(i, i, a[i]));
  st.insert(node(n + 1, n + 1, 0));
  long long oup = 0;
  while (Q--) {
    long long opt = getrand() % 4 + 1, x, l = getrand() % n + 1,
              r = getrand() % n + 1;
    if (r < l) swap(l, r);
    if (opt == 3)
      x = getrand() % (r - l + 1) + 1;
    else
      x = getrand() % vmax + 1;
    if (opt == 4) mod = getrand() % vmax + 1;
    if (opt == 1)
      add(l, r, x);
    else if (opt == 2)
      assign(l, r, x);
    else if (opt == 3)
      oup++, printf("%lld\n", rank(l, r, x));
    else
      oup++, printf("%lld\n", query(l, r, x));
  }
  return 0;
}
}  // namespace yspm
signed main() { return yspm::main(); }