codeforces#P896C. Willem, Chtholly and Seniorious

Willem, Chtholly and Seniorious

Description

— Willem...

— What's the matter?

— It seems that there's something wrong with Seniorious...

— I'll have a look...

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. .

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.

For each operation of types 3 or 4, output a line containing the answer.

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.

Samples

10 10 7 9

2
1
0
3

10 10 9 9

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