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:

def func(x,y):
    return min(x,y)

class LazySegmentTree():
    def __init__(self, n):
        self.lv = (n-1).bit_length()
        self.n = 2 ** self.lv
        self.arr = [2**31-1] * (2*self.n-1)
        self.lazy = [None] * (2*self.n-1)

    #伝播区間
    def getidx(self, l, r):
        L = (l + self.n) >> 1
        R = (r + self.n) >> 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(self.lv):
            if rc <= i:
                yield R
            if L < R and lc <= i:
                yield L
            L >>= 1; R >>=1

    #伝播
    def propagate(self,*ids):
        for i in reversed(ids):
            v = self.lazy[i-1]
            if v is None:
                continue
            self.lazy[2*i-1] = self.arr[2*i-1] = self.lazy[2*i] = self.arr[2*i] = v
            self.lazy[i-1] = None

    #[l, r] をxで更新
    def update(self,l,r,x):
        #[l, r)に変更
        r+=1
        *ids, = self.getidx(l,r)
        self.propagate(*ids)

        L = self.n + l; R = self.n + r
        while L < R:
            if R & 1:
                R -=1
                self.lazy[R-1] = self.arr[R-1] = x
            if L & 1:
                self.lazy[L-1] = self.arr[L-1] = x
                L += 1
            L >>= 1; R >>= 1

        for i in ids:
            self.arr[i-1] = func(self.arr[2*i-1], self.arr[2*i])

    #[l, r]のクエリに答える
    def query(self,l,r):
        #[l, r)に変更
        r+=1
        self.propagate(*self.getidx(l,r))
        L = self.n + l; R = self.n + r

        s = 2**31-1
        while L < R:
            if R & 1:
                R -=1
                s = func(s, self.arr[R-1])
            if L & 1:
                s = func(s, self.arr[L-1])
                L += 1
            L >>= 1; R >>= 1
        return s

import sys
input=sys.stdin.readline
n,q=map(int,input().split())
sg=LazySegmentTree(n)
for _ in range(q):
    que=list(map(int,input().split()))
    if que[0]==0:
        y,l,r,x=que
        sg.update(l,r,x)
    else:
        y,l,r=que
        print(sg.query(l,r))