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 FOR(i,a,b) for(int i=(a);i<(b);++i)
#define REP(i,n)  for(int i=0;i<(n);++i)
#define FORq(i, m, n) for(int i = (m);i <= (n);++i)
#define SCD(n) scanf("%d",&n)
#define SCD2(m,n) scanf("%d%d",&m,&n)
#define SCD3(m,n,k) scanf("%d%d%d",&m,&n,&k)
#define SCLLD(n) scanf("%lld",&n)
#define SCLLD2(m,n) scanf("%lld%lld",&m,&n)
#define SCLLD3(m,n,k) scanf("%lld%lld%lld",&m,&n,&k)
#define PB push_back
#define MP make_pair
#define ARSCD(A,N) REP(i,N){SCD(A[i]);}
#define ARSCD1(A,N) FORq(i,1,N){SCD(A[i]);}
#define PRINTD(n) printf("%d\n",n)
#define PRINTLLD(n) printf("%lld\n",n)
#define DEBUG printf("%s\n","debug")
#define fst first
#define snd second
#define SIN(x,S) (S.count(x) != 0)
using namespace std;
typedef pair<int,int> PII;
typedef vector<int> VI;
typedef vector < VI > VVI;
typedef vector<long long> VL;
typedef long long ll;
typedef long long integer;
//////////////////////////////////////////////////
// Range Query - RSQ and RAQ with Lazy Segment Tree
// https://www.slideshare.net/iwiwi/ss-3578491
// http://judge.u-aizu.ac.jp/onlinejudge/description.jsp?id=DSL_2_H&lang=jp

struct StarrySkyTree{
    // for vertex n
    // par (n-1)/2
    // chi 2n+1,2n+2

    // index i
    // bottom vertex n+i-1

    int segn2; // imply actually vertex number
    vector<ll> data,lazy; 
     
    StarrySkyTree(int n){
        segn2=1;
        while(segn2<n) segn2*=2; //必要分2^Mを取る
         
        data.resize(segn2*2,0);
        lazy.resize(segn2*2,0);
    }
    
    void eval(int k,int l,int r){
        if (lazy[k] != 0){ // lazy nonzero
            data[k] += lazy[k]; // itself
            
            if (r-l > 1){
                lazy[2*k+1] += lazy[k] ;
                lazy[2*k+2] += lazy[k] ;
            }

            lazy[k] = 0;
        }
    }

    ll query(int a,int b,int l=0,int r=-1,int k=0){
        if(r<0) r=segn2; // when calling function at first
        eval(k,l,r);

        if (r <= a || b <= l) return INT_MAX; //disjoint
        if (a <= l && r <= b) return data[k]; //include
        
        ll vl = query(a,b,l,(l+r)/2,k*2+1);
        ll vr = query(a,b,(l+r)/2,r,k*2+2);
        return min(vl,vr);
    }
     
    
    void update(int a,int b ,ll x, int l = 0, int r=-1,int k = 0){
        if(r==-1) r=segn2;

        eval(k,l,r);
        
        if (b <= l || r <= a) return;

        if (a <= l && r <= b){
            lazy[k] += x;
            eval(k,l,r);
        }else{
            //一部のみ含むとき，子ノードを計算してその結果をもらう
            update(a,b,x,l,(l+r)/2,2*k+1);
            update(a,b,x,(l+r)/2,r,2*k+2);
            data[k] = min(data[2*k+1] , data[2*k+2]);
            
        }
        
    }
};

// 0-indexed
// sum [s,t) --> s,t+1
// update [s,t,x) --> s,t+1,x

int main(){
    int N,q;
    SCD2(N,q);
    StarrySkyTree seg(N);

    REP(i,q){
        int d; SCD(d);
        if (d==0){
            int s,t,x;
            SCD3(s,t,x);
            seg.update(s,t+1,x);
        }else{
            int s,t; SCD2(s,t);
            
            PRINTLLD(seg.query(s,t+1));
        }
    }
}