Write a program which manipulates a sequence $A$ = {$a_0, a_1, ..., a_{n-1}$} with the following operations:

* $add(s, t, x)$ : add $x$ to $a_s, a_{s+1}, ..., a_t$.
* $find(s, t)$ : report the minimum value in $a_s, a_{s+1}, ..., a_t$.



Note that the initial values of $a_i ( i = 0, 1, ..., n-1 )$ are 0.

Constraints

* $1 ≤ n ≤ 100000$
* $1 ≤ q ≤ 100000$
* $0 ≤ s ≤ t < n$
* $-1000 ≤ x ≤ 1000$

Input


$n$ $q$
$query_1$
$query_2$
:
$query_q$


In the first line, $n$ (the number of elements in $A$) and $q$ (the number of queries) are given. Then, $i$th query $query_i$ 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 $add(s, t, x)$ and '1' denotes $find(s, t)$.

Output

For each $find$ query, print the minimum value.

Example

Input

6 7
0 1 3 1
0 2 4 -2
1 0 5
1 0 1
0 3 5 3
1 3 4
1 0 5


Output

-2
0
1
-1
#include <bits/stdc++.h>
using namespace std;

static const int MAX_SIZE = 1<<18;
static const long LLong = 1<<30;

typedef long long Int;
Int segMin[2 * MAX_SIZE - 1], segAdd[2 * MAX_SIZE - 1];

void add(int a, int b, Int x, int k = 0, int l = 0, int r = MAX_SIZE){
    
    if(r <= a || b <= l) return;
    
    if(a <= l && r <= b){
        segAdd[k] += x;
        return;
    }
    
    add(a, b, x, k * 2 + 1, l, (l + r) / 2);
    add(a, b, x, k * 2 + 2, (l + r) / 2, r);
    
    segMin[k] = min(segMin[k * 2 + 1] + segAdd[k * 2 + 1], segMin[k * 2 + 2] + segAdd[k * 2 + 2]);
}

Int getMin(int a, int b, int k = 0, int l = 0, int r = MAX_SIZE){
    
    if(r <= a || b <= l) return LLong;
    
    if(a <= l && r <= b) return (segMin[k] + segAdd[k]);
    
    Int left = getMin(a, b, k * 2 + 1, l, (l + r) / 2);
    Int right = getMin(a, b, k * 2 + 2, (l + r) / 2, r);
    
    return min(left, right) + segAdd[k];
}

int main(){
    for(int i = 0; i < 2 * MAX_SIZE - 1; i++)
        segMin[i] = segAdd[i] = 0;
    int n, q, b, s, t, x;
    cin >> n >> q;
    for(int i = 0; i < q; i++){
        cin >> b >> s >> t;
        if(b) cout << getMin(s, t + 1) << endl;
        else {
            cin >> x;
            add(s, t + 1, x);
        }
    }
    return 0;
}
