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 oo = 0x3f3f3f3f;
const long long ooo = 9223372036854775807ll;
const int _cnt = 1000 * 1000 + 7;
const int _p = 1000 * 1000 * 1000 + 7;
const int N = 501005;
const double PI = acos(-1.0);
const double eps = 1e-9;
int o(int x) { return x % _p; }
int gcd(int a, int b) { return b ? gcd(b, a % b) : a; }
int lcm(int a, int b) { return a / gcd(a, b) * b; }
void file_put() {
  freopen("filename.in", "r", stdin);
  freopen("filename.out", "w", stdout);
}
int op, l, r, n, m, vmax, a[N], s, x, y;
map<int, long long> M;
map<int, long long>::iterator L, R, it;
vector<pair<long long, int>> V;
long long ans = 0, seed;
int rnd() {
  int ret = seed;
  seed = (seed * 7 + 13) % _p;
  return ret;
}
long long Pow(long long x, long long k, long long p) {
  x %= p;
  if (k == 0) return 1;
  if (k == 1) return x;
  long long t = Pow(x, k >> 1, p);
  (t *= t) %= p;
  if (k & 1)
    return t * x % p;
  else
    return t;
}
int main() {
  scanf("%d%d%I64d%d", &n, &m, &seed, &vmax);
  for (int i = (1); i <= (n); ++i) a[i] = rnd() % vmax + 1, M[i] = a[i];
  M[n + 1] = 0;
  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;
    L = M.upper_bound(l);
    --L;
    if (L->first != l) M[l] = L->second, ++L;
    R = M.upper_bound(r + 1);
    --R;
    if (R->first != r + 1) M[r + 1] = R->second, ++R;
    if (op == 1)
      while (L != R) L->second += x, ++L;
    if (op == 2) {
      while (L != R) L = M.erase(L);
      M[l] = x;
    }
    if (op == 3) {
      V.clear();
      while (L != R)
        it = L, it++, V.push_back(make_pair(L->second, it->first - L->first)),
        ++L;
      sort((V).begin(), (V).end());
      s = 0;
      for (__typeof((V).begin()) it = (V).begin(); it != (V).end(); it++) {
        s += it->second;
        if (s >= x) {
          printf("%I64d\n", it->first);
          break;
        }
      }
    }
    if (op == 4) {
      ans = 0;
      while (L != R)
        it = L, it++,
        (ans += Pow(L->second, x, y) * (it->first - L->first) % y) %= y, ++L;
      printf("%I64d\n", ans);
    }
  }
  return 0;
}