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 <iostream>
#include <cstdio>
#include <algorithm>
#include <cmath>
#include <vector>
#include <list>
#include <set>
#include <map>
#include <queue>
#include <stack>
#include <cctype>
#include <cassert>
#include <climits>
#include <string>
#include <bitset>
#include <cfloat>
//#include <unordered_set>

using namespace std;

#define int long long

//sqrtN -> 最大値のみに対応できればいいので，それを使う
const int sqrtN = 512;
struct SqrtDecomposition {
  int N, K;
  vector<int> data;
  vector<int> bucketData;
  vector<int> bucketMin;
  const int INIT_VALUE = 2147483647;
  
  SqrtDecomposition(int n) : N(n) {
    //分割数
    K = (N + sqrtN - 1) / sqrtN;
    data.assign(K * sqrtN, INIT_VALUE);
    bucketData.assign(K, -1);
    bucketMin.assign(K, INIT_VALUE);
  }
  //ここまで共通
  
  void eval(int k) {
    //バケツを分配
    int l = k * sqrtN;
    int r = (k + 1) * sqrtN;
    if(bucketData[k] != -1) {
      for(int i = l; i < r; ++i) {
        data[i] = bucketData[k];
      }
      bucketData[k] = -1;
    }
  }
  void update(int s, int t, int x) {
    for(int k = 0; k < K; ++k) {
      //各バケツをループ
      int l = k * sqrtN;
      int r = (k + 1) * sqrtN;
      if(s >= r || t <= l) continue;
      if(l >= s && t >= r) {
        //完全に区間内
        bucketData[k] = x;
        bucketMin[k] = x;
      } else {
        eval(k);
        
        for(int i = max(s, l); i < min(t, r); ++i) {
          data[i] = x;
        }
        
        int smin = INIT_VALUE;
        for(int i = l; i < r; ++i) {
          smin = min(smin, data[i]);
        }
        bucketMin[k] = smin;
      }
    }
  }
  
  int find(int s, int t) {
    int minval = INIT_VALUE;
    for(int k = 0; k < K; ++k) {
      int l = k * sqrtN;
      int r = (k + 1) * sqrtN;
      if(r <= s || l >= t) continue;
      if(r <= t && l >= s) {
        minval = min(minval, bucketMin[k]);
      } else {
        eval(k);
        for(int i = max(s, l); i < min(t, r); ++i) {
          minval = min(minval, data[i]);
        }
      }
    }
    return minval;
  }
};

signed main(void) {
  cin.tie(0); ios::sync_with_stdio(false);
  
  int n, q;
  cin >> n >> q;
  
  SqrtDecomposition sq(n);
  
  for(int i = 0; i < q; i++) {
    int qtype;
    cin >> qtype;
    if(qtype == 0) {
      int s, t, x;
      cin >> s >> t >> x;
      ++t;
      sq.update(s, t, x);
    } else {
      int s, t;
      cin >> s >> t;
      ++t;
      cout << sq.find(s, t) << endl;
    }
  }  

}