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<cstdio>
#include<algorithm>
#define N 100000
#define MAX (1<<31)-1
using namespace std;
int A[N], T[4*N], lazy[4*N];

void build(int l, int r, int k) {
    if (l == r) {
        T[k] = A[l];
        return;
    }
    int mid = (l + r) / 2;
    build(l, mid, k * 2);
    build(mid + 1, r, k * 2 + 1);
    T[k] = min(T[2 * k], T[2 * k + 1]);
}
void pushdown(int k) {
    if (lazy[k] != -1) {
        lazy[2*k] = lazy[k];
        lazy[2*k+1] = lazy[k];
        T[2*k] =  lazy[k];
        T[2*k+1] = lazy[k];
        lazy[k] = -1;
    }
}
void updata(int L, int R, int data ,int l, int r ,int k) {
    if (L <= l && r <= R) {
        T[k] = data;
        lazy[k] = data;
        return;
    }
    pushdown(k);
    int mid = (l + r) / 2;
    if (mid >= L) updata(L, R, data, l, mid, 2*k);
    if (mid < R) updata(L, R, data, mid+1, r, 2*k+1);
    T[k] = min(T[2 * k], T[2 * k + 1]);
}
int find(int L, int R, int l, int r, int k) {
    if (L <= l && r <= R) return T[k];
    pushdown(k);
    int mid = (l + r) / 2;
    int ans = MAX;
    if (mid >= L) ans = min(ans, find(L, R, l, mid, k * 2));
    if (mid < R) ans = min(ans, find(L, R, mid + 1, r, k * 2 + 1));
    return ans;
}
int main() {
    int n, q;
    scanf("%d%d", &n, &q);
    for (int i = 0; i < n; i++)
        A[i] = MAX;
    for (int i = 0; i < 4*n; i++)
        lazy[i] = -1;
    build(0, n - 1, 1);
    int com, s, t, x, y;
    while (q--) {
        scanf("%d", &com);
        if (com == 0) {
            scanf("%d%d%d", &s, &t, &x);
            updata(s, t, x, 0, n - 1, 1);
        }
        else {
            scanf("%d%d", &s, &t);
            printf("%d\n", find(s, t, 0, n - 1, 1));
        }
    }
    return 0;
}