Problem: 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

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
#define rep(i,n) for(int i = 0; i < (n); ++i)
#define srep(i,s,t) for (int i = s; i < t; ++i)
#define drep(i,n) for(int i = (n)-1; i >= 0; --i)
using namespace std;
typedef long long int ll;
typedef pair<ll,ll> P;
#define yn {puts("Yes");}else{puts("No");}
#define MAX_N 200005

long long int INF = 1e18;
int const TREE_SIZE = 1 << 21;
long long int seg_tree[TREE_SIZE] = {};
long long int c_diff[TREE_SIZE] = {}; // 子との差

// [a, b) の最小値を求める
// find(a, b + 1, 1, 0, 0, TREE_SIZE / 2)
long long int find(int a, int b, int index, long long int diff, int l, int r){
	seg_tree[index] += diff;
	c_diff[index] += diff;
	
	if(r <= a || b <= l){
		return INF;
	}
	if(a <= l && r <= b){
		return seg_tree[index];
	}
	
	long long int ret = INF;
	ret = min(ret, find(a, b, index * 2, c_diff[index], l, (l + r) / 2));
	ret = min(ret, find(a, b, index * 2 + 1, c_diff[index], (l + r) / 2, r));
	c_diff[index] = 0;
	return ret;
}

// [a, b) に num を足す
// add(a, b + 1, 1, num, 0, 0, TREE_SIZE / 2);
long long int add(int a, int b, int index, long long int num, long long int diff, int l, int r){
	seg_tree[index] += diff;
	c_diff[index] += diff;
	
	if(r <= a || b <= l){
		return seg_tree[index];
	}
	if(a <= l && r <= b){
		seg_tree[index] += num;
		c_diff[index] += num;
		return seg_tree[index];
	}
	
	long long int ret = INF;
	ret = min(ret, add(a, b, index * 2, num, c_diff[index], l, (l + r) / 2));
	ret = min(ret, add(a, b, index * 2 + 1, num, c_diff[index], (l + r) / 2, r));
	c_diff[index] = 0;
	return seg_tree[index] = ret;
}

int main(){
	
	int n, q;
	cin >> n >> q;
	
	for(int loop = 0; loop < q; loop++){
		long long int num, s, t, x;
		cin >> num;
		if(num == 0){
			cin >> s >> t >> x;
			add(s, t + 1, 1, x, 0, 0, TREE_SIZE / 2);
		}else{
			cin >> s >> t;
			cout << find(s, t + 1, 1, 0, 0, TREE_SIZE / 2) << endl;
		}
	}
	
	return 0;
}