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;
long long seed, vmax, a[100005], t[100005];
struct info {
  int l, r;
  long long val;
  bool operator<(const info &p) const { return r < p.r; }
} A[100005], B[100005];
struct node {
  long long val;
  int size;
  bool operator<(const node &p) const { return val < p.val; }
} q[100005];
multiset<info> s;
multiset<info>::iterator it;
long long Pow(long long a, long long b, long long mod) {
  long long res = 1;
  a %= mod;
  while (b) {
    if (b & 1) res = res * a % mod;
    a = a * a % mod;
    b >>= 1;
  }
  return res;
}
long long rnd() {
  long long ret = seed;
  seed = (seed * 7 + 13) % 1000000007;
  return ret;
}
int main() {
  int op, l, r, x, y;
  scanf("%d%d%I64d%I64d", &n, &m, &seed, &vmax);
  for (int i = 1; i <= n; i++)
    a[i] = (rnd() % vmax) + 1, s.insert((info){i, i, a[i]});
  s.insert((info){n + 1, 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;
    it = s.lower_bound((info){l, l, 0});
    if (op == 1) {
      int l1 = 0, l2 = 0;
      for (; (*it).l <= r; ++it) {
        info tmp = *it;
        B[++l2] = tmp;
        if (tmp.l < l && r < tmp.r) {
          A[++l1] = (info){tmp.l, l - 1, tmp.val};
          A[++l1] = (info){l, r, tmp.val + x};
          A[++l1] = (info){r + 1, tmp.r, tmp.val};
          continue;
        }
        if (tmp.l < l && tmp.r >= l) {
          A[++l1] = (info){tmp.l, l - 1, tmp.val};
          A[++l1] = (info){l, tmp.r, tmp.val + x};
          continue;
        }
        if (tmp.l <= r && tmp.r > r) {
          A[++l1] = (info){tmp.l, r, tmp.val + x};
          A[++l1] = (info){r + 1, tmp.r, tmp.val};
          continue;
        }
        A[++l1] = (info){tmp.l, tmp.r, tmp.val + x};
      }
      for (int i = 1; i <= l2; i++) s.erase(B[i]);
      for (int i = 1; i <= l1; i++) s.insert(A[i]);
    }
    if (op == 2) {
      int l1 = 0, l2 = 0;
      for (; (*it).l <= r; ++it) {
        info tmp = *it;
        B[++l2] = tmp;
        if (tmp.l < l && tmp.r >= l) A[++l1] = (info){tmp.l, l - 1, tmp.val};
        if (tmp.l <= r && tmp.r > r) A[++l1] = (info){r + 1, tmp.r, tmp.val};
      }
      A[++l1] = (info){l, r, x};
      for (int i = 1; i <= l2; i++) s.erase(B[i]);
      for (int i = 1; i <= l1; i++) s.insert(A[i]);
    }
    if (op == 3) {
      int tot = 0, sum = 0;
      for (; (*it).l <= r; ++it)
        q[++tot] = (node){(*it).val, min((*it).r, r) - max((*it).l, l) + 1};
      sort(q + 1, q + tot + 1);
      for (int i = 1; i <= tot; sum += q[i].size, i++)
        if (sum + q[i].size >= x) {
          printf("%I64d\n", q[i].val);
          break;
        }
    }
    if (op == 4) {
      int ans = 0;
      for (; (*it).l <= r; ++it) {
        int num = min((*it).r, r) - max((*it).l, l) + 1;
        ans += (long long)Pow((*it).val, x, y) * num % y;
        if (ans >= y) ans -= y;
      }
      printf("%d\n", ans);
    }
  }
  return 0;
}