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 int N = 1e5 + 5, P = 1e9 + 7;
inline void write(long long x) {
  if (x > 9) write(x / 10ll);
  putchar(x % 10ll + '0');
}
struct miaow {
  int l, r;
  mutable long long sum;
  miaow(int l = 0, int r = 0, long long sum = 0) : l(l), r(r), sum(sum) {}
  inline bool operator<(const miaow& qwe) const { return l < qwe.l; }
};
int n, m;
set<miaow> kdl;
set<miaow>::iterator fg(int x) {
  set<miaow>::iterator qwe = kdl.lower_bound(miaow(x, 0, 0));
  if (qwe != kdl.end() && qwe->l == x) return qwe;
  --qwe;
  int l = qwe->l, r = qwe->r;
  long long sum = qwe->sum;
  kdl.erase(qwe);
  kdl.insert(miaow(l, x - 1, sum));
  return kdl.insert(miaow(x, r, sum)).first;
}
void xg(int l, int r, long long sum) {
  set<miaow>::iterator rrr = fg(r + 1), lll = fg(l);
  kdl.erase(lll, rrr);
  kdl.insert(miaow(l, r, sum));
}
void add(int l, int r, long long sum) {
  set<miaow>::iterator rrr = fg(r + 1), lll = fg(l);
  for (set<miaow>::iterator i = lll; i != rrr; ++i) {
    i->sum += sum;
  }
}
long long qpow(long long qwe, long long asd, long long p) {
  long long sum = 1;
  qwe %= p;
  while (asd) {
    if (asd & 1) sum = sum * qwe % p;
    qwe = qwe * qwe % p;
    asd >>= 1ll;
  }
  return sum;
}
long long cf(int l, int r, long long x, long long y) {
  long long ans = 0;
  set<miaow>::iterator rrr = fg(r + 1), lll = fg(l);
  for (set<miaow>::iterator i = lll; i != rrr; ++i) {
    long long gs = ((i->r) - (i->l) + 1), sum = i->sum;
    ans = (ans + qpow(sum, x, y) * gs) % y;
  }
  return ans;
}
struct meow {
  long long a;
  int b;
  meow(long long a = 0, int b = 0) : a(a), b(b) {}
  inline bool operator<(const meow& qwe) const { return a < qwe.a; }
};
meow dl[N];
int dlh = 0;
long long pm(int l, int r, int x) {
  dlh = 0;
  set<miaow>::iterator rrr = fg(r + 1), lll = fg(l);
  for (set<miaow>::iterator i = lll; i != rrr; ++i) {
    dl[dlh++] = meow(i->sum, (i->r) - (i->l) + 1);
  }
  sort(dl, dl + dlh);
  for (int i = 0; i < dlh; ++i) {
    x -= dl[i].b;
    if (x <= 0) return dl[i].a;
  }
}
void sc() {
  for (set<miaow>::iterator i = kdl.begin(); i != kdl.end(); ++i) {
    int gs = i->r - i->l + 1, sum = i->sum;
    for (int j = 0; j < gs; ++j) {
      cout << sum << " ";
    }
  }
  cout << endl;
}
long long seed, umax;
long long rnd() {
  long long sum = seed;
  seed = (seed * 7 + 13) % P;
  return sum;
}
void csh() {
  cin >> n >> m >> seed >> umax;
  for (int i = 1; i <= n; ++i) {
    long long x = rnd() % umax + 1;
    kdl.insert(miaow(i, i, x));
  }
}
void xw() {
  while (m) {
    --m;
    int op = rnd() % 4 + 1, l = rnd() % n + 1, r = rnd() % n + 1;
    if (l > r) swap(l, r);
    long long x, y, ans;
    if (op == 3)
      x = (rnd() % (r - l + 1)) + 1;
    else
      x = (rnd() % umax) + 1;
    if (op == 4) y = rnd() % umax + 1;
    switch (op) {
      case 1:
        add(l, r, x);
        break;
      case 2:
        xg(l, r, x);
        break;
      case 3:
        ans = pm(l, r, x);
        write(ans), putchar('\n');
        break;
      case 4:
        ans = cf(l, r, x, y);
        write(ans), putchar('\n');
        break;
    }
  }
}
int main() {
  csh();
  xw();
  return 0;
}