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 rep(i,n) for(int i = 0; i < (n);i++)
#define sz(x) int(x.size())
typedef long long ll;
typedef pair<int,int> P;
const ll INF = (1LL<<60);
/*RMQ & RUQ*/
template <typename T> struct LazySegmentTree {
  
  int n;
  vector<T> node, lazy;

  LazySegmentTree (int sz, T init = 0) {
    n = 1; while (n < sz) n <<= 1;
    node.assign(2*n, init);
    lazy.assign(2*n,-1); 
  }

  void eval (int k) {
    if (lazy[k] == -1) return ;
    node[k] = lazy[k];
    if (k < n) {
      lazy[2*k+1] = lazy[k];
      lazy[2*k+2] = lazy[k];
    }
    lazy[k] = -1;
  }

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

  T query(int a, int b, int k = 0, int l = 0, int r = -1) {
    eval(k);
    if (r < 0) r = n;
    if (r <= a || b <= l) return INF;
    if (a <= l && r <= b) return node[k];
    T vl = query(a, b, 2*k+1, l, (l+r)/2);
    T vr = query(a, b, 2*k+2, (l+r)/2, r);
    return min(vl, vr);
  }

};

int main(){ 
  int n, q;
  cin >> n >> q;
  LazySegmentTree<ll> sg(n, (1LL<<31)-1);
  while (q--) {
    int c, a, b;
    ll x;
    cin >> c >> a >> b;
    if (c) {
      cout << sg.query(a,b+1) << endl;
    } else {
      cin >> x;
      sg.update(a,b+1,x);
    }
  }
  return 0;
}