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;

int n, q;
int segSize = 1;
int tree[300000], lazy[300000] = {};
bool isLazy[300000];

void init(){
  while(segSize < n) segSize *= 2;

  for(int i = 0;i < segSize;i++){
    tree[segSize-1+i] = INT_MAX;
  }

  for(int i = segSize-2;i >= 0;i--){
    tree[i] = min(tree[i*2+1], tree[i*2+2]);
  }
}

void eval(int k, int l, int r){

  if(isLazy[k]){
    tree[k] = lazy[k];

    if(r - l > 1){
      isLazy[2*k+1] = isLazy[2*k+2] = true;
      lazy[2*k+1] = lazy[2*k+2] = lazy[k];
    }

    isLazy[k] = false;
  }

  return;
}

void update(int a, int b, int x, int k = 0, int l = 0, int r = -1){
  if(r == -1) r = segSize;

  eval(k, l, r);

  if(r <= a || b <= l) return;

  if(a <= l && r <= b){
    isLazy[k] = true;
    lazy[k] = x;
    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);
    tree[k] = min(tree[2*k+1],tree[2*k+2]);
  }
}

int getmin(int a, int b, int k = 0, int l = 0, int r = -1){
  if(r == -1) r= segSize;

  eval(k, l, r);

  if(r <= a || b <= l) return INT_MAX;
  if(a <= l && r <= b) return tree[k];

  return min(getmin(a, b, 2*k+1, l, (l+r)/2), getmin(a, b, 2*k+2, (l+r)/2, r));
}

int main(){
  cin >> n >> q;
  init();

  int p, s, t, x;
  for(int i = 0;i < q;i++){
    cin >> p;
    if(p == 0){
      cin >> s >> t >> x;
      update(s, t+1, x);
    }else{
      cin >> s >> t;
      cout << getmin(s, t+1) << endl;
    }
  }

  return 0;
}