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 MAXN = 100010;
long long seed;
int a[MAXN], le[4 * MAXN], ri[4 * MAXN];
long long data[4 * MAXN];
map<long long, int> s;
int rnd() {
  int ret = seed;
  seed = (seed * 7 + 13) % 1000000007;
  return ret;
}
void init(int p, int l, int r) {
  le[p] = l;
  ri[p] = r;
  if (l == r) {
    data[p] = a[l];
    return;
  }
  data[p] = -1;
  int m = (l + r) / 2;
  init(p * 2, l, m);
  init(p * 2 + 1, m + 1, r);
}
void add(int p, int l, int r, int x) {
  if (x == 0) return;
  if (data[p] != -1) {
    if (le[p] >= l && ri[p] <= r) {
      data[p] += x;
      return;
    }
    data[p * 2] = data[p];
    data[p * 2 + 1] = data[p];
    data[p] = -1;
  }
  int m = (le[p] + ri[p]) / 2;
  if (m >= l) add(p * 2, l, r, x);
  if (m + 1 <= r) add(p * 2 + 1, l, r, x);
}
void assign(int p, int l, int r, int x) {
  if (le[p] >= l && ri[p] <= r) {
    data[p] = x;
    return;
  }
  if (data[p] != -1 && data[p] != x) {
    data[p * 2] = data[p];
    data[p * 2 + 1] = data[p];
    data[p] = -1;
  }
  int m = (le[p] + ri[p]) / 2;
  if (m >= l) assign(p * 2, l, r, x);
  if (m + 1 <= r) assign(p * 2 + 1, l, r, x);
}
void getrank(int p, int l, int r) {
  if (data[p] != -1) {
    int len = min(r, ri[p]) - max(l, le[p]) + 1;
    s[data[p]] += len;
    return;
  }
  int m = (le[p] + ri[p]) / 2;
  if (m >= l) getrank(p * 2, l, r);
  if (m + 1 <= r) getrank(p * 2 + 1, l, r);
}
long long getpow(long long b, int c, int y) {
  long long ans = 1, cur = b % y;
  while (c > 0) {
    if (c % 2 == 1) ans = (ans * cur) % y;
    c /= 2;
    cur = (cur * cur) % y;
  }
  return ans;
}
long long sum(int p, int l, int r, int x, int y) {
  long long tot = 0;
  if (data[p] != -1) {
    tot = (min(r, ri[p]) - max(l, le[p]) + 1) * getpow(data[p], x, y) % y;
    return tot;
  }
  int m = (le[p] + ri[p]) / 2;
  if (m >= l) tot = (tot + sum(p * 2, l, r, x, y)) % y;
  if (m + 1 <= r) tot = (tot + sum(p * 2 + 1, l, r, x, y)) % y;
  return tot;
}
int main() {
  int n, m, vmax;
  cin >> n >> m >> seed >> vmax;
  for (int i = 1; i <= n; i++) a[i] = (rnd() % vmax) + 1;
  init(1, 1, n);
  int nout = 0;
  while (m--) {
    int op = (rnd() % 4) + 1;
    int l = (rnd() % n) + 1;
    int r = (rnd() % n) + 1;
    int x, y = 0;
    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;
    switch (op) {
      case 1:
        add(1, l, r, x);
        break;
      case 2:
        assign(1, l, r, x);
        break;
      case 3: {
        s.erase(s.begin(), s.end());
        getrank(1, l, r);
        int tmp = 0;
        nout++;
        for (map<long long, int>::iterator it = s.begin(); it != s.end();
             it++) {
          tmp += it->second;
          if (tmp >= x) {
            cout << it->first << endl;
            break;
          }
        }
        break;
      }
      case 4: {
        nout++;
        cout << sum(1, l, r, x, y) << endl;
        break;
      }
    }
  }
}