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 = 100005;
int n, m, op, seed, vmax, tot, l, r, ret;
long long c1, c2, a[N], x, y;
struct data {
  long long x;
  int len;
} dat[N];
set<int> s;
set<int>::iterator p;
int getint() {
  char ch;
  int f = 1;
  while (!isdigit(ch = getchar()))
    if (ch == '-') f = -1;
  int x = ch - 48;
  while (isdigit(ch = getchar())) x = x * 10 + ch - 48;
  return x * f;
}
int rnd() {
  ret = seed;
  seed = (seed * 7ll + 13) % 1000000007;
  return ret;
}
bool cmp(data a, data b) { return a.x < b.x; }
long long getmi(long long a, int x) {
  long long rt = 1;
  a %= y;
  while (x) {
    if (x & 1) rt = rt * a % y;
    a = a * a % y, x >>= 1;
  }
  return rt;
}
int main() {
  n = getint(), m = getint(), seed = getint(), vmax = getint();
  for (int i = 1; i <= n; ++i) a[i] = rnd() % vmax + 1;
  for (int i = 0; i <= n + 1; ++i) s.insert(i);
  while (m--) {
    op = rnd() % 4 + 1, l = rnd() % n + 1, r = rnd() % n + 1;
    if (l > r) swap(l, r);
    if (op == 3)
      x = rnd() % (r - l + 1) + 1;
    else
      x = rnd() % vmax + 1;
    if (op == 4) y = rnd() % vmax + 1;
    if (op == 1) {
      p = --s.upper_bound(r), c2 = a[*p];
      p = s.lower_bound(r + 1);
      if (*p != r + 1) s.insert(r + 1), a[r + 1] = c2;
      p = --s.upper_bound(l), c1 = a[*p];
      if (*p != l) s.insert(l);
      a[l] = c1 + x;
      for (p = s.upper_bound(l); *p <= r; ++p) a[*p] += x;
    }
    if (op == 2) {
      p = --s.upper_bound(r), c2 = a[*p];
      p = s.lower_bound(r + 1);
      if (*p != r + 1) s.insert(r + 1), a[r + 1] = c2;
      p = --s.upper_bound(l), c1 = a[*p];
      if (*p != l) s.insert(l);
      a[l] = x;
      for (p = s.upper_bound(l); *p <= r;) s.erase(*p), p = s.upper_bound(l);
    }
    if (op == 3 || op == 4) {
      p = --s.upper_bound(l), dat[tot = 1] = (data){a[*p], l};
      for (++p; *p <= r; ++p) dat[++tot] = (data){a[*p], *p};
      for (int i = 1; i <= tot; ++i)
        dat[i].len =
            i == tot ? r - dat[i].len + 1 : dat[i + 1].len - dat[i].len;
      if (op == 3) {
        sort(dat + 1, dat + 1 + tot, cmp);
        for (int i = 1; i <= tot; ++i) {
          if (x <= dat[i].len) {
            printf("%I64d\n", dat[i].x);
            break;
          }
          x -= dat[i].len;
        }
      } else {
        long long ans = 0;
        for (int i = 1; i <= tot; ++i)
          ans = (ans + getmi(dat[i].x, x) * dat[i].len) % y;
        printf("%I64d\n", ans);
      }
    }
  }
  return 0;
}