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;

template<typename T> class LazySegmentTree {
    int n, sz;
    vector<T> dat, Lazy;
    const T INF = (1 << 30) - 1;

    void eval(int k){//ノードkの値を更新して遅延配列を伝播
        dat[k] += Lazy[k];//ココ
        if(Lazy[k]){
            if(k < n - 1) {
                Lazy[2 * k + 1] += Lazy[k];
                Lazy[2 * k + 2] += Lazy[k];
            }
            Lazy[k] = 0;
        }
    }

public:
    LazySegmentTree(const vector<T> &v) : sz((int)v.size()) {
        //nは元配列の要素数以上で最小の二冪
        //datは[n-1, 2n-1]が元の配列に対応　最下段より上段の要素数がn-1
        n = 1;
        while(n < sz)n *= 2;
        dat.resize(2 * n - 1, INF);//ノードを単位元で初期化
        Lazy.resize(2 * n - 1, 0);//遅延配列を0で初期化

        for(int i = 0; i < sz; i++){
            dat[i + n - 1] = v[i];//元配列に対応するノードを初期化
        }
        for(int i = n - 2; i >= 0; i--){
            dat[i] = min(dat[2 * i + 1], dat[2 * i + 2]);//下から上になめて更新
        }
    }

    //[a,b)にxを一様加算
    //今見てるノードのインデックス:k  それに対応する元配列の区間[l,r)
    //ノードの値を更新

    void add(int a, int b, T x, int k = 0, int l = 0, int r = -1){
        if(r < 0)r = n;
        eval(k);//ココ
        if(r <= a || b <= l){//更新区間にふくまれない
            return;
        }
        if(a <= l && r <= b){//更新区間に完全に含まれる
            Lazy[k] += x;
            eval(k);
        }
        else {//一部更新区間を含む
            add(a, b, x, 2 * k + 1, l, (l + r) / 2);
            add(a, b, x, 2 * k + 2, (l + r) / 2, r);
            if(k < n - 1){//ココ
                dat[k] = min(dat[2 * k + 1], dat[2 * k + 2]);
            }
        }
        
    }

    //[a,b)のminを求める
    T query(int a, int b, int k = 0, int l = 0, int r = -1){
        if(r < 0)r = n;
        if(r <= a || b <= l){
            return INF;//目的の区間を含まないので単位元を返す
        }

        eval(k);
        if(a <= l && r <= b){//遅延配列を更新してノードの値を返す
            return dat[k];
        }
        else {//左右に配って帰ってきた二つのminを返す
            T vl = query(a, b, 2 * k + 1, l, (l + r) / 2);
            T vr = query(a, b, 2 * k + 2, (l + r) / 2, r);
            return min(vl, vr);
        }
    }
};

int main(){
    vector<int> V;
    int N, Q;
    cin >> N >> Q;
    V.resize(N, 0);
    LazySegmentTree<int> seg(V);
    for(int i = 0; i < Q; i++){
        int a, s, t, x;
        cin >> a >> s >> t;
        if(a == 0){
            cin >> x;
            seg.add(s, t + 1, x);
        }
        else {
            cout << seg.query(s, t + 1) << endl;
        }
    }
    

    return 0;
}