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 <vector>
#include <cmath>

using namespace std;

const int DEFAULT = 2147483647;

//Regular lazy propagation RMQ
struct SegmentTree {
    int n;
    vector<int> heap;
    vector<int> lazy;
    
    SegmentTree(int n): n(n) {
        heap.assign(4 * n, DEFAULT);
        lazy.assign(4 * n, DEFAULT);
    }
    
    void update(int nodeId, int l, int r, int ql, int qr, int v) {
        if (qr < l || ql > r) {
            return;
        }
        else if (ql <= l && r <= qr) {
            lazy[nodeId] = v;
        }
        else {
            pushdown(nodeId);
            int m = l + (r - l)/2;
            update(nodeId * 2, l, m, ql, qr, v);
            update(nodeId * 2 + 1, m + 1, r, ql, qr, v);
            pushup(nodeId);
        }
    }
    
    void pushdown(int nodeId) {
        if (lazy[nodeId] != -1) {
            lazy[nodeId * 2] = lazy[nodeId];
            lazy[nodeId * 2 + 1] = lazy[nodeId];
            heap[nodeId] = lazy[nodeId];
            lazy[nodeId] = -1;
        }
    }
      
      void pushup(int nodeId) {
          heap[nodeId] = min(getMin(nodeId * 2), getMin(nodeId * 2 + 1));
      }
      
      int getMin(int nodeId) {
          if (lazy[nodeId] != -1) {
              return lazy[nodeId];
          }
          else {
              return heap[nodeId];
          }
      }
    
    int query(int nodeId, int l, int r, int ql, int qr) {
        if (qr < l || ql > r) {
            return DEFAULT;
        }
        if (ql <= l && r <= qr) {
            return getMin(nodeId);
        } 
        else {
            pushdown(nodeId);
            int m = l + (r - l)/2;
            return min(query(nodeId * 2, l, m, ql, qr),
                    query(nodeId * 2 + 1, m + 1, r, ql, qr));
        }
    }  
};

int main() {
    int n, q;
    cin >> n >> q;
    SegmentTree st(n);
    for (int i = 0; i < q; i++) {
        int cmd;
        cin >> cmd;
        if (cmd == 0) {
            int ql, qr, v;
            cin >> ql >> qr >> v;
            st.update(1, 0, n - 1, ql, qr, v);
//            for (int i = 0; i < 4 * n; i++) {
//                cout << i << ":" << st.heap[i] << endl; 
//            }
        }
        else {
            int ql, qr;
            cin >> ql >> qr;
            cout << st.query(1, 0, n - 1, ql, qr) << endl;
        }
    }
    
    return 0;
}