Problem: Write a program which manipulates a sequence A = {a0, a1, . . . , an−1} with the following operations:

* update(s, t, x): change as, as+1, ..., at to x.
* find(s, t): report the minimum element in as, as+1, ..., at.



Note that the initial values of ai (i = 0, 1, . . . , n−1) are 231-1.

Constraints

* 1 ≤ n ≤ 100000
* 1 ≤ q ≤ 100000
* 0 ≤ s ≤ t < n
* 0 ≤ x < 231−1

Input


n q
query1
query2
:
queryq


In the first line, n (the number of elements in A) and q (the number of queries) are given. Then, ith query queryi is given in the following format:


0 s t x


or


1 s t


The first digit represents the type of the query. '0' denotes update(s, t, x) and '1' denotes find(s, t).

Output

For each find operation, print the minimum value.

Examples

Input

3 5
0 0 1 1
0 1 2 3
0 2 2 2
1 0 2
1 1 2


Output

1
2


Input

1 3
1 0 0
0 0 0 5
1 0 0


Output

2147483647
5

Solution (in PYTHON3):

Solution:
import sys
readline = sys.stdin.readline
write = sys.stdout.write
 
N, Q = map(int, input().split())
INF = 2**31-1
 
LV = (N-1).bit_length()
N0 = 2**LV
data = [INF]*(2*N0)
lazy = [None]*(2*N0)
 
def gindex(l, r):
    L = (l + N0) >> 1; R = (r + N0) >> 1
    lc = 0 if l & 1 else (L & -L).bit_length()
    rc = 0 if r & 1 else (R & -R).bit_length()
    for i in range(LV):
        if rc <= i:
            yield R
        if L < R and lc <= i:
            yield L
        L >>= 1; R >>= 1
 
def propagates(*ids):
    for i in reversed(ids):
        v = lazy[i-1]
        if v is None:
            continue
        lazy[2*i-1] = data[2*i-1] = lazy[2*i] = data[2*i] = v
        lazy[i-1] = None
 
def update(l, r, x):
    *ids, = gindex(l, r)
    propagates(*ids)
 
    L = N0 + l; R = N0 + r
    while L < R:
        if R & 1:
            R -= 1
            lazy[R-1] = data[R-1] = x
        if L & 1:
            lazy[L-1] = data[L-1] = x
            L += 1
        L >>= 1; R >>= 1
    for i in ids:
        data[i-1] = min(data[2*i-1], data[2*i])
 
def query(l, r):
    propagates(*gindex(l, r))
    L = N0 + l; R = N0 + r
 
    s = INF
    while L < R:
        if R & 1:
            R -= 1
            s = min(s, data[R-1])
        if L & 1:
            s = min(s, data[L-1])
            L += 1
        L >>= 1; R >>= 1
    return s
 
 
answer = []
for q in range(Q):
    t, *cmd = map(int, readline().split())
    if t:
        s, t = cmd
        answer.append(str(query(s, t+1)))
    else:
        s, t, x = cmd
        update(s, t+1, x)
 
write("\n".join(answer))
write("\n")