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 CPP):

Solution:
#include<bits/stdc++.h>

using namespace std;

const int N=1e5+10;

typedef vector<int> vec;

class RUQ
{
public:
    vec data;
    vec lazy;
    int n;
    const int INF=(1LL<<31)-1;

    RUQ(int _n)
    {
        data.resize(4*_n);
        lazy.resize(4*_n);
        for(n=1;n<_n;n<<=1);
        fill(data.begin(),data.end(),(1LL<<31)-1);
        fill(lazy.begin(),lazy.end(),-1);
    }

    void Down(int p,int l,int r)
    {
        if(lazy[p]==-1) return;
        data[p]=lazy[p];
        if(l+1<r)
        {
            lazy[p*2+1]=lazy[p];
            lazy[p*2+2]=lazy[p];
        }
        lazy[p]=-1;
    }

    int update(int p,int l,int r,int x,int a,int b)
    {
        Down(p,l,r);
        if(a>=r||b<=l) return data[p];
        if(a<=l&&r<=b)
        {
            lazy[p]=x;
            return data[p]=lazy[p];
        }
        int mid=(l+r)/2;
        return data[p]=min(update(p*2+1,l,mid,x,a,b),update(p*2+2,mid,r,x,a,b));
    }

    int query(int p,int l,int r,int a,int b)
    {
        if(a>=r||b<=l) return INF;
        Down(p,l,r);
        if(a<=l&&r<=b)
        {
            return data[p];
        }
        int mid=(l+r)/2;
        return min(query(p*2+1,l,mid,a,b),query(p*2+2,mid,r,a,b));
    }
};

int main()
{
    int n,q;

    while(scanf("%d %d",&n,&q)!=EOF)
    {

        RUQ ruq(n);
        while(q--)
        {
            int op,a,b,c;
            scanf("%d",&op);
            if(op)
            {
                scanf("%d %d",&a,&b);
                printf("%d\n",ruq.query(0,0,ruq.n,a,b+1));
            }else
            {
                scanf("%d %d %d",&a,&b,&c);
                ruq.update(0,0,ruq.n,c,a,b+1);
            }
        }
    }

    return 0;
}