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
#include <cmath>
#include <cstdio>
#include <vector>
#include <iostream>
#include <algorithm>
#include <queue>
#include <string>
#include <set>
#include <map>
#define REP(i,n) for(ll i = 0; i < (ll)n; i++)
#define INF 1000000000000000
using namespace std;
typedef long long ll;
typedef double db;
typedef string str;

const ll MAX_N = 1000000;
ll segAdd[MAX_N*2-1], segMin[MAX_N*2-1];
ll size = 1;

void init(ll n){
  while(size<n) size*=2;
  REP(i,2*size-1) segAdd[i] = 0, segMin[i] = 0;
}

void add(ll a, ll b, ll x, ll k, ll l, ll r){
  if(r<=a||b<=l) return;
  if(a<=l&&r<=b){
    segAdd[k]+=x;
    return;
  }
  add(a,b,x,k*2+1,l,(l+r)/2);
  add(a,b,x,k*2+2,(l+r)/2,r);

  segMin[k] = min(segMin[k*2+1]+segAdd[k*2+1],segMin[k*2+2]+segAdd[k*2+2]);
}

ll getMin(ll a, ll b, ll k, ll l, ll r){
  if(r<=a||b<=l) return INF;
  if(a<=l&&r<=b) return segMin[k]+segAdd[k];

  ll left = getMin(a,b,k*2+1,l,(l+r)/2);
  ll right = getMin(a,b,k*2+2,(l+r)/2,r);
  return min(left,right)+segAdd[k];
}

int main(){
  ll n,q;
  cin >> n >> q;
  init(n);
  vector<ll> ans;
  REP(_,q){
    int i; cin >> i;
    if(!i){
      ll s,t,x; cin >> s >> t >> x;
      add(s,t+1,x,0,0,size);
    }else{
      ll s,t; cin >> s >> t;
      ans.push_back(getMin(s,t+1,0,0,size));
    }
  }
  for(ll a: ans) cout << a << endl;
  return 0;
}

