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>

using namespace std;

template<typename T>
class LazySegTree {
public:
    explicit LazySegTree(int n, T init, T def) : N(calcN_(n)), def(def) {
        mVal.assign(2*N+1, init);
        mLazy.assign(2*N+1, 0);
        mDirty.assign(2*N+1, 0);
    }
    void update(int l, int r, T value){
        updateImpl_(l, r, value, 0, 0, N);
    }
    T get(int l, int r){
        return getImpl_(l, r, 0, 0, N);
    }
private:
    int calcN_(int n){
        int res = 1;
        while(res < n) res *= 2;
        return res;
    }
    void updateImpl_(int l, int r, T value, int idx, int rangeL, int rangeR){
        eval_(idx, rangeL, rangeR);
        if(r <= rangeL || rangeR <= l) return;
        if(l <= rangeL && rangeR <= r){
            setLazy_(idx, value);
            eval_(idx, rangeL, rangeR);
        } else {
            int rangeM = (rangeL+rangeR)/2;
            updateImpl_(l, r, value, 2*idx+1, rangeL, rangeM);
            updateImpl_(l, r, value, 2*idx+2, rangeM, rangeR);
            mVal[idx] = operate_(mVal[2*idx+1], mVal[2*idx+2]);
        }
    }
    void setLazy_(int idx, T value){
        mLazy[idx] = value;
        mDirty[idx] = 1;
    }
    T getImpl_(int l, int r, int idx, int rangeL, int rangeR){
        if(r <= rangeL || rangeR <= l) return def;
        eval_(idx, rangeL, rangeR);
        if(l <= rangeL && rangeR <= r) return mVal[idx];
        int rangeM = (rangeL+rangeR)/2;
        T a = getImpl_(l, r, 2*idx+1, rangeL, rangeM);
        T b = getImpl_(l, r, 2*idx+2, rangeM, rangeR);
        return operate_(a, b);
    }
    void eval_(int idx, int rangeL, int rangeR){
        if(!mDirty[idx]) return;
        mVal[idx] = mLazy[idx];
        if(idx < N-1){
            setLazy_(2*idx+1, mLazy[idx]);
            setLazy_(2*idx+2, mLazy[idx]);
        }
        mLazy[idx] = 0;
        mDirty[idx] = 0;
    }
    T operate_(T a, T b) const { return min(a, b); }
    const int N;
    const T def;
    vector<T> mVal; 
    vector<T> mLazy;
    vector<int> mDirty;
};

int main(){
    int n, q;
    while(cin >> n >> q){
        LazySegTree<int> seg(n, 0x7FFFFFFF, 0x7FFFFFFF);
        for(int i=0;i<q;i++){
            int c; cin >> c;
            if(c == 0){
                int s, t, x; cin >> s >> t >> x;
                seg.update(s, t+1, x);
            } else {
                int s, t; cin >> s >> t;
                cout << seg.get(s, t+1) << endl;
            }
        }
    }
}