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;
template <typename T>
ostream& operator<<(ostream& os, const vector<T>& v) {
  for (auto i : v) os << i << " ";
  return os;
}
template <typename T>
ostream& operator<<(ostream& os, const set<T>& v) {
  for (auto i : v) os << i << " ";
  return os;
}
template <typename T>
T read() {
  T sgn = 1, res = 0;
  char c = getchar();
  while (c < '0' || c > '9') {
    if (c == '-') sgn = -sgn;
    c = getchar();
  }
  while ('0' <= c && c <= '9') {
    res = res * 10 + c - '0';
    c = getchar();
  }
  return res * sgn;
}
class node {
 public:
  mutable long long l, r, val;
  node(long long l = 0, long long r = 0, long long val = 0)
      : l(l), r(r), val(val) {}
  bool operator<(const node& cmp) const {
    if (l == cmp.l) return r < cmp.r;
    return l < cmp.l;
  }
};
class heapnode {
 public:
  long long len, val;
  heapnode(long long len = 0, long long val = 0) : len(len), val(val) {}
  bool operator<(const heapnode& cmp) const {
    if (val == cmp.val) return len < cmp.len;
    return val < cmp.val;
  }
};
const long long MOD = 1e9 + 7;
const int N = 1e5 + 233;
long long dig[N], seed, n, m, op, vm, l, r, x, y;
set<node> s;
vector<heapnode> v;
long long rnd() {
  long long ret = seed;
  seed = (seed * 7 + 13) % MOD;
  return ret;
}
void getData() {
  scanf("%d%d%d%d", &n, &m, &seed, &vm);
  for (int i = 1; i <= n; ++i) dig[i] = (rnd() % vm) + 1;
}
void getOp() {
  op = (rnd() % 4) + 1;
  l = (rnd() % n) + 1;
  r = (rnd() % n) + 1;
  x = y = 0;
  if (l > r) swap(l, r);
  if (op == 3)
    x = (rnd() % (r - l + 1)) + 1;
  else
    x = (rnd() % vm) + 1;
  if (op == 4) y = (rnd() % vm) + 1;
}
set<node>::iterator split(int pos) {
  auto p = s.lower_bound(node(pos));
  if (p != s.end() && p->l == pos) return p;
  p--;
  if (pos > p->r) return s.end();
  long long l = p->l;
  long long r = p->r;
  long long v = p->val;
  s.erase(p);
  s.emplace(l, pos - 1, v);
  return s.emplace(pos, r, v).first;
}
long long fp(long long x, long long y, long long mod) {
  long long res = 1;
  x %= mod;
  while (y) {
    if (y & 1) res *= x;
    x *= x;
    x %= mod;
    res %= mod;
    y >>= 1;
  }
  return res;
}
int main() {
  getData();
  for (int i = 1; i <= n; ++i) s.emplace(i, i, dig[i]);
  while (m--) {
    getOp();
    split(l);
    auto sr = split(r + 1);
    auto sl = split(l);
    switch (op) {
      case 1: {
        for (; sl != sr; ++sl) sl->val += x;
        break;
      }
      case 2: {
        s.erase(sl, sr);
        s.emplace(l, r, x);
        break;
      }
      case 3: {
        vector<heapnode>().swap(v);
        bool ok = false;
        for (; sl != sr; ++sl) v.emplace_back(sl->r - sl->l + 1, sl->val);
        sort(v.begin(), v.end());
        for (auto i : v) {
          x -= i.len;
          if (x <= 0) {
            printf("%I64d\n", i.val);
            ok = true;
            break;
          }
        }
        assert(ok);
        break;
      }
      case 4: {
        long long res = 0;
        while (sl != sr) {
          long long temp = (sl->r - sl->l + 1) * fp(sl->val, x, y);
          temp %= y;
          res += temp;
          res %= y;
          sl++;
        }
        printf("%I64d\n", res);
        break;
      }
      default:
        assert(false);
    }
  }
  return 0;
}