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;
struct node {
  int l, r;
  long long x;
  node() {}
  node(int l_, int r_, long long x_) {
    l = l_;
    r = r_;
    x = x_;
  }
  bool operator<(const node &t) const { return l < t.l; }
} tmp[100010];
pair<long long, int> a[100010];
int n, m, seed, vmax;
set<node> st;
int rnd() {
  int ret = seed;
  seed = (seed * 7LL + 13) % 1000000007;
  return ret;
}
long long power(long long x, long long n, long long mod) {
  x %= mod;
  long long ans = 1;
  while (n) {
    if (n & 1) ans = ans * x % mod;
    x = x * x % mod;
    n >>= 1;
  }
  return ans;
}
int main() {
  scanf("%d%d%d%d", &n, &m, &seed, &vmax);
  for (int i = 1; i <= n; ++i) st.insert(node(i, i, (rnd() % vmax) + 1));
  while (m--) {
    int op = (rnd() % 4) + 1;
    int l = rnd() % n + 1, r = rnd() % n + 1, x, y;
    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) {
      auto it = st.upper_bound(node(l, l, 0));
      --it;
      int cnt = 0;
      for (; it != st.end() && it->l <= r;) {
        if (it->l < l) tmp[cnt++] = node(it->l, l - 1, it->x);
        tmp[cnt++] = node(max(it->l, l), min(r, it->r), it->x + x);
        if (r < it->r) tmp[cnt++] = node(r + 1, it->r, it->x);
        it = st.erase(it);
      }
      for (int i = 0; i < cnt; ++i) st.insert(tmp[i]);
    } else if (op == 2) {
      auto it = st.upper_bound(node(l, l, 0));
      --it;
      int cnt = 0;
      tmp[cnt++] = node(l, r, x);
      for (; it != st.end() && it->l <= r;) {
        if (it->l < l) tmp[cnt++] = node(it->l, l - 1, it->x);
        if (r < it->r) tmp[cnt++] = node(r + 1, it->r, it->x);
        it = st.erase(it);
      }
      for (int i = 0; i < cnt; ++i) st.insert(tmp[i]);
    } else if (op == 3) {
      auto it = st.upper_bound(node(l, l, 0));
      --it;
      int cnt = 0;
      for (; it != st.end() && it->l <= r; ++it) {
        a[cnt++] = make_pair(it->x, min(r, it->r) - max(it->l, l) + 1);
      }
      sort(a, a + cnt);
      int sum = 0;
      for (int i = 0; i < cnt; ++i) {
        sum += a[i].second;
        if (x <= sum) {
          printf("%lld\n", a[i].first);
          break;
        }
      }
    } else {
      auto it = st.upper_bound(node(l, l, 0));
      --it;
      long long ans = 0;
      for (; it != st.end() && it->l <= r; ++it)
        ans = (ans + 1LL * (min(r, it->r) - max(it->l, l) + 1) *
                         power(it->x, x, y) % y) %
              y;
      printf("%lld\n", ans);
    }
  }
}