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;
int x, y;
int Pow(int a, int anum) {
  int ret = 1;
  while (anum) {
    if (anum & 1) ret = 1ll * ret * a % y;
    a = 1ll * a * a % y;
    anum >>= 1;
  }
  return ret;
}
int ql, qr, a[(100000 + 5)];
namespace Seg_tree {
long long h[(100000 + 5) << 2], mx[(100000 + 5) << 2], mn[(100000 + 5) << 2];
bool tag[(100000 + 5) << 2];
void Pushdown(int o) {
  if (!tag[o]) return;
  tag[o] = false;
  tag[(o << 1)] = tag[(o << 1 | 1)] = true;
  h[(o << 1)] = h[(o << 1 | 1)] = mx[(o << 1)] = mx[(o << 1 | 1)] =
      mn[(o << 1)] = mn[(o << 1 | 1)] = h[o];
}
void Pushup(int o) {
  mx[o] = max(mx[(o << 1)], mx[(o << 1 | 1)]);
  mn[o] = min(mn[(o << 1)], mn[(o << 1 | 1)]);
  if (!tag[(o << 1)] || !tag[(o << 1 | 1)] || h[(o << 1)] != h[(o << 1 | 1)]) {
    tag[o] = false;
    return;
  }
  h[o] = h[(o << 1)];
  tag[o] = true;
}
void Build(int o, int L, int R) {
  if (L == R) {
    tag[o] = true;
    h[o] = mx[o] = mn[o] = a[L];
    return;
  }
  Build((o << 1), L, ((L + R) >> 1));
  Build((o << 1 | 1), ((L + R) >> 1) + 1, R);
  Pushup(o);
}
void Modify1(int o, int L, int R) {
  if (ql <= L && qr >= R) {
    tag[o] = true;
    h[o] = mx[o] = mn[o] = x;
    return;
  }
  Pushdown(o);
  if (ql <= ((L + R) >> 1)) Modify1((o << 1), L, ((L + R) >> 1));
  if (qr > ((L + R) >> 1)) Modify1((o << 1 | 1), ((L + R) >> 1) + 1, R);
  Pushup(o);
}
void Modify2(int o, int L, int R) {
  if (ql <= L && qr >= R && tag[o]) {
    h[o] += x;
    mn[o] += x;
    mx[o] += x;
    return;
  }
  Pushdown(o);
  if (ql <= ((L + R) >> 1)) Modify2((o << 1), L, ((L + R) >> 1));
  if (qr > ((L + R) >> 1)) Modify2((o << 1 | 1), ((L + R) >> 1) + 1, R);
  Pushup(o);
}
int Query1(int o, int L, int R, long long av) {
  if (av < mn[o]) return 0;
  if (tag[o]) {
    int l = max(L, ql), r = min(R, qr);
    return r - l + 1;
  }
  int ret = 0;
  if (ql <= ((L + R) >> 1)) ret += Query1((o << 1), L, ((L + R) >> 1), av);
  if (qr > ((L + R) >> 1))
    ret += Query1((o << 1 | 1), ((L + R) >> 1) + 1, R, av);
  return ret;
}
int Query2(int o, int L, int R) {
  if (tag[o]) {
    int l = max(L, ql), r = min(R, qr);
    return 1ll * Pow(h[o] % y, x) * (r - l + 1) % y;
  }
  int ret = 0;
  if (ql <= ((L + R) >> 1)) ret += Query2((o << 1), L, ((L + R) >> 1));
  if (qr > ((L + R) >> 1)) ret += Query2((o << 1 | 1), ((L + R) >> 1) + 1, R);
  return ret % y;
}
};  // namespace Seg_tree
using namespace Seg_tree;
int n, m, vmax;
long long seed;
void Solve(int k) {
  long long L = 1, R = 1ll * vmax * (m + 1), mid;
  while (L < R) {
    mid = (L + R) >> 1;
    if (Query1(1, 1, n, mid) >= k)
      R = mid;
    else
      L = mid + 1;
  }
  printf("%I64d\n", L);
}
int rnd() {
  long long ret = seed;
  seed = (seed * 7 + 13) % 1000000007;
  return ret;
}
int main() {
  int op;
  scanf("%d%d%I64d%d", &n, &m, &seed, &vmax);
  for (int i = (1); i <= (int)(n); ++i) a[i] = (rnd() % vmax) + 1;
  Build(1, 1, n);
  for (int i = (1); i <= (int)(m); ++i) {
    op = (rnd() % 4) + 1;
    ql = (rnd() % n) + 1;
    qr = (rnd() % n) + 1;
    if (ql > qr) swap(ql, qr);
    if (op == 3)
      x = (rnd() % (qr - ql + 1)) + 1;
    else
      x = (rnd() % vmax) + 1;
    if (op == 4) y = (rnd() % vmax) + 1;
    if (op == 1)
      Modify2(1, 1, n);
    else if (op == 2)
      Modify1(1, 1, n);
    else if (op == 3)
      Solve(x);
    else
      printf("%d\n", Query2(1, 1, n));
  }
  return 0;
}