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 n, m, seed, vmax;
long long arr[100000], upd[340];
int rnd() {
  int ret = seed;
  seed = (seed * 7LL + 13) % 1000000007;
  return ret;
}
void update_block(int ibl) {
  long long x = upd[ibl] > 0 ? upd[ibl] : -upd[ibl];
  if (upd[ibl] > 0)
    for (int i = ibl * 340, ii = min(i + 340, n); i < ii; i++) arr[i] += x;
  else if (upd[ibl] < 0)
    for (int i = ibl * 340, ii = min(i + 340, n); i < ii; i++) arr[i] = x;
  upd[ibl] = 0;
}
void alter_count(map<long long, int> &mp, long long k, int i) {
  if (mp.count(k))
    mp[k] += i;
  else
    mp[k] = i;
}
long long power(long long x, int n, int y) {
  long long r = (n & 1) ? x : 1, rr = n == 1 ? 1 : power(x, n >> 1, y);
  return r * rr % y * rr % y;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> m >> seed >> vmax;
  for (int i = 0; i < n; i++) arr[i] = rnd() % vmax + 1;
  for (int iq = 0; iq < m; iq++) {
    int op = rnd() % 4;
    int l = rnd() % n, r = rnd() % n, x, y = 0;
    if (l > r) swap(l, r);
    r++;
    if (op == 2)
      x = rnd() % (r - l) + 1;
    else
      x = rnd() % vmax + 1;
    if (op == 3) y = rnd() % vmax + 1;
    if (op == 0) {
      if (l / 340 == (r - 1) / 340) {
        update_block(l / 340);
        for (int i = l; i < r; i++) arr[i] += x;
      } else {
        if (l % 340 != 0) {
          update_block(l / 340);
          for (int i = l, ii = min(l / 340 * 340 + 340, n); i < ii; i++)
            arr[i] += x;
        }
        if (r % 340 != 0) {
          update_block(r / 340);
          for (int i = r / 340 * 340; i < r; i++) arr[i] += x;
        }
        for (int i = l % 340 == 0 ? l / 340 : l / 340 + 1, j = r / 340; i < j;
             i++)
          if (upd[i] >= 0)
            upd[i] += x;
          else
            upd[i] = -(-upd[i] + x);
      }
    } else if (op == 1) {
      if (l / 340 == (r - 1) / 340) {
        update_block(l / 340);
        for (int i = l; i < r; i++) arr[i] = x;
      } else {
        if (l % 340 != 0) {
          update_block(l / 340);
          for (int i = l, ii = min(l / 340 * 340 + 340, n); i < ii; i++)
            arr[i] = x;
        }
        if (r % 340 != 0) {
          update_block(r / 340);
          for (int i = r / 340 * 340; i < r; i++) arr[i] = x;
        }
        for (int i = l % 340 == 0 ? l / 340 : l / 340 + 1, j = r / 340; i < j;
             i++)
          upd[i] = -x;
      }
    } else if (op == 2 || op == 3) {
      map<long long, int> mp;
      if (l / 340 == (r - 1) / 340) {
        if (upd[l / 340] >= 0)
          for (int i = l; i < r; i++) alter_count(mp, arr[i] + upd[l / 340], 1);
        else
          alter_count(mp, -upd[l / 340], r - l);
      } else {
        if (l % 340 != 0) {
          if (upd[l / 340] >= 0)
            for (int i = l, ii = min(l / 340 * 340 + 340, n); i < ii; i++)
              alter_count(mp, arr[i] + upd[l / 340], 1);
          else
            alter_count(mp, -upd[l / 340], min(l / 340 * 340 + 340, n) - l);
        }
        if (r % 340 != 0) {
          if (upd[r / 340] >= 0)
            for (int i = r / 340 * 340; i < r; i++)
              alter_count(mp, arr[i] + upd[r / 340], 1);
          else
            alter_count(mp, -upd[r / 340], r - r / 340 * 340);
        }
        for (int i = l % 340 == 0 ? l / 340 : l / 340 + 1, j = r / 340; i < j;
             i++)
          if (upd[i] >= 0)
            for (int k = i * 340, kk = min(i * 340 + 340, n); k < kk; k++)
              alter_count(mp, arr[k] + upd[i], 1);
          else
            alter_count(mp, -upd[i], min(i * 340 + 340, n) - i * 340);
      }
      long long ans = 0;
      if (op == 2) {
        ans = -1;
        for (auto &r : mp) {
          if (r.second >= x) {
            ans = r.first;
            break;
          } else
            x -= r.second;
        }
      } else {
        for (auto &r : mp)
          ans = (ans + power(r.first % y, x, y) * r.second % y) % y;
      }
      cout << ans << endl;
    }
  }
  return 0;
}