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 vector<int> vec;

class Seg
{
public:
    int n;
    vec data;
    vec lazy;
    const int INF=1e9+10;

    Seg(int _n)
    {
        data.resize(4*_n);
        lazy.resize(4*_n);
        for(n=1;n<_n;n<<=1);
    }

    void Down(int p,int l,int r)
    {
        data[p]+=lazy[p];
        if(l+1<r)
        {
            lazy[p*2+1]+=lazy[p];
            lazy[p*2+2]+=lazy[p];
        }
        lazy[p]=0;
    }

    int add(int p,int l,int r,int x,int a,int b)
    {
        if(a>=r||b<=l) return data[p]+lazy[p];
        else if(a<=l&&r<=b)
        {
            lazy[p]+=x;
            return data[p]+lazy[p];
        }else
        {
            Down(p,l,r);
            int mid=(l+r)/2;
            return data[p]=min(add(p*2+1,l,mid,x,a,b),add(p*2+2,mid,r,x,a,b));
        }
    }

    int query(int p,int l,int r,int a,int b)
    {
        if(a>=r||b<=l) return INF;
        else if(a<=l&&r<=b)
        {
            return data[p]+lazy[p];
        }else
        {
            Down(p,l,r);
            int mid=(l+r)/2;
            return min(query(p*2+1,l,mid,a,b),query(p*2+2,mid,r,a,b));
        }
    }
};

int main()
{
    int n,q;

    while(scanf("%d %d",&n,&q)!=EOF)
    {
        Seg seg(n);
        while(q--)
        {
            int op,a,b,c;
            scanf("%d",&op);

            if(op)
            {
                scanf("%d %d",&a,&b);
                printf("%d\n",seg.query(0,0,seg.n,a,b+1));
            }else
            {
                scanf("%d %d %d",&a,&b,&c);
                seg.add(0,0,seg.n,c,a,b+1);
            }
        }
    }

    return 0;
}