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;
#define int long long

const int INF = (1LL << 31) - 1;
vector<int> v;

struct DelaySegmentTree {
  int n;
  vector<int> node,delay;
  vector<bool> delayflag;

  // 元配列 v をセグメント木で表現する
  void init(){
    n = 1; while(n < v.size()) n *= 2;
    node.resize(2*n-1,INF);
    delay.resize(2*n-1,INF);
    delayflag.resize(2*n-1,false);

    for(int i = 0; i < v.size(); i++) node[i+n-1] = v[i];
    for(int i = n-2; i >= 0; i--) node[i] = min(node[2*i+1],node[2*i+2]);
  }

  void eval(int k,int l,int r){
    if(delayflag[k]){
      node[k] = delay[k];
      if(r-l > 1){
        delay[2*k+1] = delay[k];
        delayflag[2*k+1] = true;
        delay[2*k+2] = delay[k];
        delayflag[2*k+2] = true;
      }
      delayflag[k] = false;
    }
  }

  void update(int a,int b,int x){ update(a,b,x,0,0,n); }
  void update(int a,int b,int x,int k,int l,int r){
    eval(k,l,r);
    if(b <= l || r <= a) return;
    if(a <= l && r <= b){
      delay[k] = x;
      delayflag[k] = true;
      eval(k,l,r);
    }
    else {
      update(a,b,x,2*k+1,l,(l+r)/2);
      update(a,b,x,2*k+2,(l+r)/2,r);
      node[k] = min(node[2*k+1],node[2*k+2]);
    }
  }

  int getmin(int a,int b){ return getmin(a,b,0,0,n); }
  int getmin(int a,int b,int k,int l,int r){
    //区間 [a, b) 中の要素の最小値を返す
    eval(k,l,r);
    if(b <= l || r <= a) return INF;
    if(a <= l && r <= b) return node[k];
    int vl = getmin(a, b, 2*k+1, l, (l+r)/2);
    int vr = getmin(a, b, 2*k+2, (l+r)/2, r);
    return min(vl, vr);
  }
}seg;

signed main(){
  cin.tie(0);
  ios::sync_with_stdio(false);

  int n,q; cin >> n >> q;
  v.resize(n,INF);
  seg.init();
  for(int i = 0; i < q; i++){
    int com,s,t,x; cin >> com;
    if(com){
      cin >> s >> t;
      cout << seg.getmin(s,t+1) << endl;
    }
    else {
      cin >> s >> t >> x;
      seg.update(s,t+1,x);
    }
  }

  return 0;
}