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; }

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

/*
 update : [s,t) に xを加算
 query  : [s,t) の　最大(小)値を出力する
 */

/* [0..N-1] */
const ll INIT = 0; // 問題に合わせる
struct SegTree {
    int N;
    ll init_v;
    vector<ll> node, lazy;
    
    SegTree(int _N):init_v(INIT) {
        N = 1;
        while (N < _N) N *= 2;
        node.resize(2 * N - 1, init_v);
        lazy.resize(2 * N - 1, init_v);
    }
    
    void lazy_evaluate(int l, int r,int k){
        node[k] += lazy[k];
        if(r - l > 1){
            lazy[2 * k + 1] += lazy[k];
            lazy[2 * k + 2] += lazy[k];
        }
        lazy[k] = 0;
    }
    
    /* [a,b) 引数の範囲に注意!! s~tまでを更新→update(s,t+1,~) */
    void update(int a, int b, ll x) { update(a, b, 0, 0, N, x); }
    void update(int a, int b, int k, int l, int r, ll x) {
        lazy_evaluate(l,r,k);
        if (r <= a || b <= l) return;
        if (a <= l && r <= b) {
            lazy[k] += x;
            lazy_evaluate(l,r,k);
        }
        else {
            update(a, b, 2 * k + 1, l, (l + r) / 2, x);
            update(a, b, 2 * k + 2, (l + r) / 2, r, x);
            node[k] = min(node[2 * k + 1],node[2 * k + 2]);
        }
    }
    
    /* [a,b) 引数の範囲に注意!! */
    ll query(int a, int b) { return query(a, b, 0, 0, N); }
    ll query(int a, int b, int k, int l, int r) {
        lazy_evaluate(l, r, k);
        if (r <= a || b <= l) return LLONG_MAX; // min : LLONG_MAX , max : LLONG_MIN
        if (a <= l && r <= b) {
            return node[k];
        }
        else {
            ll vl = query(a, b, 2 * k + 1, l, (l + r) / 2);
            ll vr = query(a, b, 2 * k + 2, (l + r) / 2, r);
            return min(vl, vr);
        }
    }
};

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);
        }else{
            int s,t; cin >> s >> t;
            cout << ST.query(s,t+1) <<endl;
        }
    }
    
    return 0;
}