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>
using namespace std;
typedef long long ll;
typedef pair<ll, ll> P;
const ll MOD = 1e9+7;
const ll INF = 1e18;
#define rep(i,m,n) for(ll i = (m); i < (n); i++)
#define rrep(i,m,n) for(ll i = (m); i >= (n); i--)
#define print(x) cout << (x) << endl;
#define printa(x,m,n) for(ll i = (m); i <= n; i++){cout << (x[i]) << " ";} cout<<endl;

template<typename T> class segtree {
private:
    int n,sz,h;
    vector<pair<T, int> > node;
    vector<T> lazy;
    void eval(int k) {
        if(lazy[k]){
            node[k].first += lazy[k];
            if(k < n) {
                lazy[k*2] += lazy[k], lazy[k*2+1] += lazy[k];
            }
            lazy[k] = 0;
        }
    }
 
public:
    segtree(const vector<T>& v) : sz((int)v.size()), h(0) {
        n = 1;
        while(n < sz) n *= 2, h++;
        node.resize(2*n, pair<T, int>(numeric_limits<T>::max(), sz));
        lazy.resize(2*n, 0);
        for(int i = 0; i < sz; i++){
            node[i+n] = make_pair(v[i], i);
        }
        for(int i = n-1; i >= 1; i--){
             node[i] = min(node[2*i], node[2*i+1]);
        }
    }
    void range(int a, int b, T x, int k=1, int l=0, int r=-1){
        if(r < 0) r = n;
        eval(k);
        if(b <= l || r <= a){
            return;
        }
        if(a <= l && r <= b){
            lazy[k] += x;
            eval(k);
        }else{
            range(a, b, x, 2*k, l, (l+r)/2);
            range(a, b, x, 2*k+1, (l+r)/2, r);
            node[k] = min(node[2*k], node[2*k+1]);
        }
    }
    pair<T, int> query(int a, int b) {
        a += n, b += n - 1;
        for(int i = h; i > 0; i--) eval(a >> i), eval(b >> i);
        b++;
        pair<T, int> res1 = make_pair(numeric_limits<T>::max(), sz);
        pair<T, int> res2 = make_pair(numeric_limits<T>::max(), sz);
        while(a < b) {
            if(a & 1) eval(a), res1 = min(res1, node[a++]);
            if(b & 1) eval(--b), res2 = min(res2, node[b]);
            a >>= 1, b >>= 1;
        }
        return min(res1, res2);
    }
};

int main(){
	cin.tie(0); ios::sync_with_stdio(false);
	
	ll n, q;
	cin >> n >> q;
	vector<ll> v(n, 0);
	segtree<ll> sg(v);
	
	rep(i, 0, q){
		ll f;
		cin >> f;
		if(f == 0){
			ll s, t, x;
			cin >> s >> t >> x;
			sg.range(s, t+1, x);
		}
		else{
			ll s, t;
			cin >> s >> t;
			print(sg.query(s, t+1).first)
		}
	}
	return 0;
}