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;
long long a[100005];
int n, m, seed, vmax;
set<int> s;
pair<long long, int> b[100005];
int rnd() {
  int tmp = seed;
  seed = (1ll * seed * 7 + 13) % 1000000007;
  return tmp;
}
int power(int x, int y, int p) {
  int s = 1 % p;
  for (; y; y /= 2, x = 1ll * x * x % p)
    if (y & 1) s = 1ll * s * x % p;
  return s;
}
int main() {
  scanf("%d%d%d%d", &n, &m, &seed, &vmax);
  for (int i = 1; i <= n; i++) a[i] = rnd() % vmax + 1, s.insert(i);
  s.insert(n + 1);
  s.insert(0);
  for (int i = 1; i <= m; i++) {
    int op = rnd() % 4 + 1;
    int l = rnd() % n + 1;
    int r = rnd() % n + 1;
    if (l > r) swap(l, r);
    int x, y;
    set<int>::iterator it;
    if (op == 3)
      x = rnd() % (r - l + 1) + 1;
    else
      x = rnd() % vmax + 1;
    if (op == 4) y = rnd() % vmax + 1;
    if (op == 1) {
      it = --s.upper_bound(r);
      long long lav2 = a[*it];
      it = --s.upper_bound(l);
      long long lav1 = a[*it];
      if (*it != l) s.insert(l);
      a[l] = lav1 + x;
      it = s.upper_bound(r);
      if (*it != r + 1) s.insert(r + 1), a[r + 1] = lav2;
      it = s.upper_bound(l);
      for (; *it <= r; a[*it++] += x)
        ;
    } else if (op == 2) {
      it = --s.upper_bound(r);
      long long lav2 = a[*it];
      it = --s.upper_bound(l);
      if (*it != l) s.insert(l);
      a[l] = x;
      it = s.upper_bound(r);
      if (*it != r + 1) s.insert(r + 1), a[r + 1] = lav2;
      it = s.upper_bound(l);
      for (; *it <= r;) {
        a[*it] = x;
        s.erase(it);
        it = s.upper_bound(l);
      }
    } else {
      it = --s.upper_bound(l);
      int top = 1;
      b[1].second = l;
      b[1].first = a[*it];
      for (it++; *it <= r; it++) {
        b[++top].second = *it;
        b[top].first = a[*it];
      }
      int lav = r + 1;
      for (int i = top; i; i--) {
        int tmp = b[i].second;
        b[i].second = lav - b[i].second;
        lav = tmp;
      }
      if (op == 3) {
        sort(b + 1, b + top + 1);
        for (int i = 1; i <= top; i++)
          if (x <= b[i].second) {
            printf("%I64d\n", b[i].first);
            break;
          } else
            x -= b[i].second;
      } else {
        int ans = 0;
        for (int i = 1; i <= top; i++)
          ans = (ans + 1ll * power(b[i].first % y, x, y) * b[i].second % y) % y;
        printf("%d\n", ans);
      }
    }
  }
}