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;

// 1-indexed
template <class T, class E>
struct SegmentTreeLaze {
  // a,b:T c,d:E e:E(unit)
  // g(f(a,b),c) = f(g(a,c),g(b,c))
  // g(g(a,c),d) = g(a,h(c,d))
  // g(a,e) = a
  typedef function<T(T, T)> F;
  typedef function<T(T, E)> G;
  typedef function<E(E, E)> H;
  int n, height;
  F f;
  G g;
  H h;
  T tunit;
  E eunit;
  vector<T> dat;
  vector<E> laz;
  SegmentTreeLaze(){};
  SegmentTreeLaze(int newn, F f, G g, H h, T nt, E ne)
      : f(f), g(g), h(h), tunit(nt), eunit(ne) {
    init(newn);
  }
  void init(int newn) {
    n = 1, height = 0;
    while(n < newn) n <<= 1, ++height;
    dat.assign(n << 1, tunit);
    laz.assign(n << 1, eunit);
  }

  inline T reflect(int k) {
    return laz[k] == eunit ? dat[k] : g(dat[k], laz[k]);
  }

  inline void eval(int k) {
    if(laz[k] == eunit) return;
    laz[k << 1] = h(laz[k << 1], laz[k]);
    laz[(k << 1) | 1] = h(laz[(k << 1) | 1], laz[k]);
    dat[k] = reflect(k);
    laz[k] = eunit;
  }

  inline void thrust(int k) {
    for(int i = height; i; --i) eval(k >> i);
  }

  void recalc(int k) {
    while(k >>= 1)
      dat[k] = f(reflect(k << 1), reflect((k << 1) | 1));
  }
  // [a,b)
  void update(int a, int b, E newdata) {
    thrust(a += n);
    thrust(b += n - 1);
    for(int l = a, r = b + 1; l < r; l >>= 1, r >>= 1) {
      if(l & 1) laz[l] = h(laz[l], newdata), l++;
      if(r & 1) --r, laz[r] = h(laz[r], newdata);
    }
    recalc(a);
    recalc(b);
  }

  void set_val(int k, T newdata) {
    thrust(k += n);
    dat[k] = newdata;
    laz[k] = eunit;
    recalc(k);
  }

  // [a,b)
  T query(int a, int b) {
    thrust(a += n);
    thrust(b += n - 1);
    T vl = tunit, vr = tunit;
    for(int l = a, r = b + 1; l < r; l >>= 1, r >>= 1) {
      if(l & 1) vl = f(vl, reflect(l++));
      if(r & 1) vr = f(reflect(--r), vr);
    }
    return f(vl, vr);
  }
};

long long n, q;

int main() {
  cin >> n >> q;
  auto f = [](long long l, long long r) {
    return min(l, r);
  };
  auto g = [](long long l, long long r) { return r; };
  auto h = [](long long l, long long r) { return r; };
  SegmentTreeLaze<long long, long long> seg(n, f, g, h,
                                            2147483647, -1);
  for(int i = 0; i < n; ++i) seg.set_val(i, 2147483647);
  for(int i = 0; i < q; ++i) {
    long long c, s, t, x;
    cin >> c >> s >> t;
    if(c) { cout << seg.query(s, t + 1) << endl; }
    else {
      cin >> x;
      seg.update(s, t + 1, x);
    }
  }
  return 0;
}