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<cstdio>
#include<algorithm>
#include<climits>
using namespace std;
static const int MAX_N = 100000;
static const int INF = INT_MAX;

int n, q;

int n_;
int dat[4 * MAX_N], laz[4 * MAX_N];

void init(){
	n_ = 1;
	while(n_ < n) n_ *= 2;
	for(int i = 0; i < 2 * n_ - 1; i++){
		dat[i] = INF;
		laz[i] = 0;
	}
	for(int i = n_ - 1; i < n_ - 1 + n; i++){
		dat[i] = 0;
		int k = i;
		while(k > 0){
			k = (k - 1) / 2;
			dat[k] = min(dat[2 * k + 1], dat[2 * k + 2]);
		}
	}
}
void add(int a, int b, int x, int k, int l, int r){
	if(r <= a || b <= l) return;
	if(a <= l && r <= b){
		laz[k] += x;
		return;
	}
	add(a, b, x, 2 * k + 1, l, (l + r) / 2);
	add(a, b, x, 2 * k + 2, (l + r) / 2, r);
	dat[k] = min(dat[2 * k + 1] + laz[2 * k + 1], dat[2 * k + 2] + laz[2 * k + 2]);
}
int find(int a, int b, int k, int l, int r){
	if(r <= a || b <= l) return INT_MAX;
	if(a <= l && r <= b) return dat[k] + laz[k];
	int vl = find(a, b, 2 * k + 1, l, (l + r) / 2);
	int vr = find(a, b, 2 * k + 2, (l + r) / 2, r);
	return min(vl, vr) + laz[k];
}

int main(){
	scanf("%d %d", &n, &q);
	init();
	for(int i = 0; i < q; i++){
		int op;
		scanf("%d", &op);
		if(op == 0){
			int s, t, x;
			scanf("%d %d %d", &s, &t, &x);
			add(s, t + 1, x, 0, 0, n_);
		}else if(op == 1){
			int s, t;
			scanf("%d %d", &s, &t);
			printf("%d\n", find(s, t + 1, 0, 0, n_));
		}
	}
	return 0;
}
