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>
struct FastIO {
  template <typename ATP>
  inline FastIO &operator>>(ATP &x) {
    x = 0;
    int f = 1;
    char c;
    for (c = getchar(); c < '0' || c > '9'; c = getchar())
      if (c == '-') f = -1;
    while (c >= '0' && c <= '9') x = x * 10 + (c ^ '0'), c = getchar();
    if (f == -1) x = -x;
    return *this;
  }
} io;
using namespace std;
template <typename ATP>
inline ATP Max(ATP x, ATP y) {
  return x > y ? x : y;
}
template <typename ATP>
inline ATP Min(ATP x, ATP y) {
  return x < y ? x : y;
}
template <typename ATP>
inline ATP Abs(ATP x) {
  return x > 0 ? x : -x;
}
const int MOD = 1e9 + 7;
const int N = 1e5 + 7;
struct nod {
  int l, r;
  mutable long long val;
  nod() {}
  nod(int _l, int _r = -1, long long _val = 0) : l(_l), r(_r), val(_val) {}
  bool operator<(const nod &rhs) const { return l < rhs.l; }
};
set<nod> S;
set<nod>::iterator Split(int pos) {
  set<nod>::iterator it = S.lower_bound(nod(pos));
  if (it != S.end() && it->l == pos) return it;
  --it;
  int r = it->r, l = it->l;
  long long val = it->val;
  S.erase(it);
  S.insert(nod(l, pos - 1, val));
  return S.insert(nod(pos, r, val)).first;
}
inline void Assign(int l, int r, long long val = 0) {
  set<nod>::iterator R = Split(r + 1), L = Split(l);
  S.erase(L, R);
  S.insert(nod(l, r, val));
}
inline void Add(int l, int r, long long val = 1) {
  set<nod>::iterator R = Split(r + 1), L = Split(l);
  for (; L != R; ++L) L->val += val;
}
inline long long Rank(int l, int r, int K) {
  vector<pair<long long, int> > V;
  set<nod>::iterator R = Split(r + 1), L = Split(l);
  V.clear();
  for (; L != R; ++L) V.push_back(make_pair(L->val, L->r - L->l + 1));
  sort(V.begin(), V.end());
  for (vector<pair<long long, int> >::iterator it = V.begin(); it != V.end();
       ++it) {
    K -= it->second;
    if (K <= 0) return it->first;
  }
  return -1ll;
}
inline long long Pow(long long a, int b, int &mod) {
  long long s = 1;
  a %= mod;
  while (b) {
    if (b & 1) s = s * a % mod;
    a = a * a % mod, b >>= 1;
  }
  return s;
}
inline long long Sum(int l, int r, int ex, int mod) {
  set<nod>::iterator R = Split(r + 1), L = Split(l);
  long long sum = 0;
  for (; L != R; ++L) {
    sum = (sum + 1ll * (L->r - L->l + 1) * Pow(L->val, ex, mod)) % mod;
  }
  return sum;
}
int n, m;
long long seed, vmax;
inline long long rnd() {
  long long ret = seed;
  seed = (seed * 7 + 13) % MOD;
  return ret;
}
long long a[N];
int main() {
  ;
  io >> n >> m >> seed >> vmax;
  for (register int i = (1); i <= (n); ++i) {
    a[i] = (rnd() % vmax) + 1;
    S.insert(nod(i, i, a[i]));
  }
  S.insert(nod(n + 1, n + 1, 0));
  for (register int i = (1); i <= (m); ++i) {
    int opt = rnd() % 4 + 1, l = rnd() % n + 1, r = rnd() % n + 1;
    if (l > r) ((l) ^= (r) ^= (l) ^= (r));
    int x, y;
    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)
      printf("%lld\n", Rank(l, r, x));
    else
      printf("%lld\n", Sum(l, r, x, y));
  }
  return 0;
}