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;

struct query{
  ll type;//0=empty, 1=add-sum ,2=set-min
  ll value;
  query(ll a=0,ll b=0):type(a),value(b) {}
};

ll INF=(1LL<<60);

struct segtree{
  ll SIZE;
  vector<query> s;
  vector<ll> t;// sum-add
  vector<ll> u;// min-set

  segtree(ll n=1){
    SIZE=1;
    while(SIZE<n)SIZE*=2;
    s.clear();
    t.clear();
    u.clear();
    s.resize(SIZE*2, query());
    t.resize(SIZE*2, 0);
    u.resize(SIZE*2, 0);
  }
  
  void func(ll k,ll l,ll r,query q){
    if(q.type==1){
      if(s[k].type==0)s[k]=q;
      else s[k].value+=q.value;
      t[k]+=q.value*(r-l);
      u[k]+=q.value;
    }
    if(q.type==2){
      s[k]=q;
      t[k]=q.value*(r-l);
      u[k]=q.value;
    }
  }

  void compute(ll k,ll l,ll r){
    query q=s[k];
    s[k]=query();
    if(q.type==0||r-l==1)return;
    ll m=(l+r)/2;
    func(k*2+1,l,m,q);
    func(k*2+2,m,r,q);
  }

  void Update(ll a,ll b,query x,ll k,ll l,ll r){
    if(b<=l || r<=a)return;
    compute(k,l,r);
    if(a<=l && r<=b){
      func(k,l,r,x);
    }else{
      ll m=(l+r)/2;
      Update(a,b,x,k*2+1,l,m);
      Update(a,b,x,k*2+2,m,r);
      t[k]=t[k*2+1]+t[k*2+2];
      u[k]=min(u[k*2+1],u[k*2+2]);
    }
  }

  ll Dfs(ll type,ll a,ll b,ll k,ll l,ll r){
    if(b<=l || r<=a){
      if(type==1)return 0; //add
      if(type==2)return INF; // min
    }
    compute(k,l,r);
    if(a<=l && r<=b){
      if(type==1)return t[k];
      if(type==2)return u[k];
    }else{
      ll m=(l+r)/2;
      ll lv=Dfs(type,a,b,k*2+1,l,m);
      ll rv=Dfs(type,a,b,k*2+2,m,r);
      if(type==1)return lv+rv; // add
      if(type==2)return min(lv,rv); // min
    }
  }

  void Add(ll a,ll b,ll x){
    Update(a,b,query(1,x), 0,0,SIZE);
  }
  void Set(ll a,ll b,ll x){
    Update(a,b,query(2,x), 0,0,SIZE);
  }
  ll Getsum(ll a,ll b){
    return Dfs(1,a,b, 0,0,SIZE);
  }

  ll Getmin(ll a,ll b){
    return Dfs(2,a,b, 0,0,SIZE);
  }
  
};



int main(){
  ll n,q;
  cin >> n >> q;
  segtree T(n);
  
  for(ll i=0;i<q;i++){
    ll type,a,b,c;
    cin >> type >> a >> b;
    if(type==0){
      cin >> c;
      T.Add(a,b+1,c);
    }
    else {
      cout << T.Getmin(a,b+1) << endl;
    }
  }
  return 0;
}