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<int, int> pii;
typedef pair<ll, ll> pll;
const int INF = 1e9;
const ll LINF = 1e18;
template<class S,class T> ostream& operator << (ostream& out,const pair<S,T>& o){ out << "(" << o.first << "," << o.second << ")"; return out; }
template<class T> ostream& operator << (ostream& out,const vector<T> V){ for(int i = 0; i < V.size(); i++){ out << V[i]; if(i!=V.size()-1) out << " ";} return out; }
template<class T> ostream& operator << (ostream& out,const vector<vector<T> > Mat){ for(int i = 0; i < Mat.size(); i++) { if(i != 0) out << endl; out << Mat[i];} return out; }
template<class S,class T> ostream& operator << (ostream& out,const map<S,T> mp){ out << "{ "; for(auto it = mp.begin(); it != mp.end(); it++){ out << it->first << ":" << it->second; if(mp.size()-1 != distance(mp.begin(),it)) out << ", "; } out << " }"; return out; }

// ================================================= //

/*starry sky tree
 add(O(logN)):区間[a, b)に値xを加算する
 getMin(O(logN)): 区間の最小値を得る
 */
static const int MAX_SIZE = 1 << 17; // 2^n !!
typedef long long ll;
ll 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; //もし交差しない区間であれば終える.
    
    if (a <= l && r <= b){ //もし今みている区間[l, r)が[a, b)に完全に内包されていれば
        segAdd[k] += x;  //区間[l, r)にkを加算する.
        return;
    }
    
    //子の区間に(必要があれば)xを加算する.
    add(a, b, x, k * 2 + 1, l, (l + r) / 2);
    add(a, b, x, k * 2 + 2, (l + r) / 2, r);
    
    //親の区間の最小値は, 子の区間の最小値 + 自分に一様に加算されている値 である.一様に加算される値は更新しなくて良い.
    segMin[k] = min(segMin[k * 2 + 1] + segAdd[k * 2 + 1], segMin[k * 2 + 2] + segAdd[k * 2 + 2]);
}

ll getMin(int a, int b, int k = 0, int l = 0, int r = MAX_SIZE){
    if (r <= a || b <= l) return (LLONG_MAX);
    if (a <= l && r <= b) return (segMin[k] + segAdd[k]); //完全に内包されていれば,その区間の最小値を返す.
    
    ll left = getMin(a, b, k * 2 + 1, l, (l + r) / 2); //子の区間の最小値を求める.
    ll right = getMin(a, b, k * 2 + 2, (l + r) / 2, r); //子の区間の最小値を求める
    
    //親の区間の最小値は, 子の区間の最小値 + 自分に一様に加算されている値
    return (min(left, right) + segAdd[k]);
}

int main(void) {
    cin.tie(0); ios::sync_with_stdio(false);
    int n,q; cin >> n >> q;
    
//    SegTree ST(n);
    while(q--){
        int com; cin >> com;
        if(com == 0){
            int s,t,x; cin >> s >> t >> x;
           // ST.update(s,t+1,x);
            add(s,t+1,x);
        }else{
            int s,t; cin >> s >> t;
            //cout << ST.query(s,t+1) <<endl;
            cout << getMin(s,t+1) << endl;
        }
    }
    
    return 0;
}