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;
const int mod = 1e9 + 7;
int rnd() {
  int rlt = seed;
  seed = (1ll * seed * 7 + 13) % mod;
  return rlt;
}
int expmod(int a, int b, int P) {
  int rlt = 1;
  while (b) {
    if (b & 1) rlt = 1ll * a * rlt % P;
    a = 1ll * a * a % P;
    b >>= 1;
  }
  return rlt;
}
long long a[101010], b[101010];
int op, le, ri, x, y;
void get_query() {
  op = (rnd() % 4) + 1;
  le = (rnd() % n) + 1;
  ri = (rnd() % n) + 1;
  if (le > ri) swap(le, ri);
  if (op == 3)
    x = (rnd() % (ri - le + 1)) + 1;
  else
    x = (rnd() % vmax) + 1;
  if (op == 4) y = (rnd() % vmax) + 1;
}
void put_query() {
  printf("%d %d %d %d", op, le, ri, x);
  if (op == 4)
    printf(" %d\n", y);
  else
    puts("");
}
long long add[101010 << 2], to[101010 << 2];
void refresh(int id) {
  if (add[id] && to[id] != -1) to[id] += add[id], add[id] = 0;
}
void push_down(int id) {
  refresh(id);
  if (to[id] != -1) {
    to[id << 1] = to[id];
    add[id << 1] = 0;
    to[(id << 1) ^ 1] = to[id];
    add[(id << 1) ^ 1] = 0;
  } else {
    add[id << 1] += add[id];
    add[(id << 1) ^ 1] += add[id];
    add[id] = 0;
  }
}
void build(int L, int R, int id) {
  if (L == R) {
    to[id] = a[L];
    return;
  }
  int mid = (L + R) >> 1;
  build(L, mid, id << 1);
  build(mid + 1, R, (id << 1) ^ 1);
}
void update(int le, int ri, int d, int L, int R, int id) {
  if (le > R || L > ri) return;
  if (le <= L && R <= ri) {
    refresh(id);
    if (to[id] != -1) {
      to[id] += d;
      return;
    }
    add[id] += d;
    return;
  }
  int mid = (L + R) >> 1;
  push_down(id);
  to[id] = -1;
  update(le, ri, d, L, mid, id << 1);
  update(le, ri, d, mid + 1, R, (id << 1) ^ 1);
}
void change(int le, int ri, int d, int L, int R, int id) {
  if (le > R || L > ri) return;
  if (le <= L && R <= ri) {
    refresh(id);
    add[id] = 0;
    to[id] = d;
    return;
  }
  int mid = (L + R) >> 1;
  push_down(id);
  to[id] = -1;
  change(le, ri, d, L, mid, id << 1);
  change(le, ri, d, mid + 1, R, (id << 1) ^ 1);
}
int query(int le, int ri, int x, int y, int L, int R, int id) {
  if (le > R || L > ri) return 0;
  if (to[id] != -1) {
    refresh(id);
    int len = min(R, ri) - max(L, le) + 1;
    return (int)((1ll * expmod(to[id] % y, x, y) * len) % y);
  }
  int mid = (L + R) >> 1;
  push_down(id);
  to[id] = -1;
  int ans1 = query(le, ri, x, y, L, mid, id << 1);
  int ans2 = query(le, ri, x, y, mid + 1, R, (id << 1) ^ 1);
  return (ans1 + ans2) % y;
}
priority_queue<pair<long long, int> > pq;
void calc(int le, int ri, int L, int R, int id) {
  if (le > R || L > ri) return;
  if (to[id] != -1) {
    refresh(id);
    int len = min(R, ri) - max(L, le) + 1;
    pq.push(make_pair(to[id], len));
    return;
  }
  int mid = (L + R) >> 1;
  push_down(id);
  to[id] = -1;
  calc(le, ri, L, mid, id << 1);
  calc(le, ri, mid + 1, R, (id << 1) ^ 1);
}
int main() {
  scanf("%d %d %d %d", &n, &m, &seed, &vmax);
  for (int i = 1; i <= n; i++) a[i] = (rnd() % vmax) + 1;
  memset(to, -1, sizeof to);
  build(1, n, 1);
  while (m--) {
    get_query();
    if (op == 1) {
      update(le, ri, x, 1, n, 1);
    } else if (op == 2) {
      change(le, ri, x, 1, n, 1);
    } else if (op == 3) {
      while (!pq.empty()) pq.pop();
      calc(le, ri, 1, n, 1);
      int tot = 0;
      x = ri - le + 1 - x;
      pair<long long, int> tmp;
      while (1) {
        tmp = pq.top();
        x -= tmp.second;
        if (x < 0) {
          printf("%I64d\n", tmp.first);
          break;
        }
        pq.pop();
      }
    } else {
      int ans = 0;
      ans = query(le, ri, x, y, 1, n, 1);
      printf("%d\n", ans);
    }
  }
  return 0;
}