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<iostream>
#include<vector>
#include<climits>
using namespace std;

// F : T x T -> T
// G : T x E -> T
// H : E x E -> E
// R : E x int(length of the range) -> E
template<typename T, typename E, typename F, typename G, typename H, typename R>
struct SegL{
private:
    vector<T> node;
    vector<E> lazy;
    int n;
    T defVal;
    E defOp;
    F f;    // merge elements(T)
    G g;    // update element(T) 
    H h;    // merge operators(E)
    R r;    // operator depends on length

public:
    SegL(int siz, T defVal, E defOp, F f, G g, H h, R r) : defVal(defVal), defOp(defOp), f(f), g(g), h(h), r(r){
        n = 1;
        while(n < siz)  n *= 2;
        node.resize(2*n-1, defVal);
        lazy.resize(2*n-1, defOp);
    }
    SegL(vector<T> v, T defVal, E defOp, F f, G g, H h, R r) : defVal(defVal), defOp(defOp), f(f), g(g), h(h), r(r){
        n = 1;
        while(n < v.size())  n *= 2;
        node.resize(2*n-1, defVal);
        lazy.resize(2*n-1, defOp);
        for(int i = 0; i < v.size(); i++)   node[n-1+i] = v[i];
        for(int i = n-2; i >= 0; i--)   node[i] = f(node[2*i+1], node[2*i+2]);
    }

    void eval(int k, int len){
        if(lazy[k] == defOp)    return;
        if(k < n-1){
            lazy[2*k+1] = h(lazy[2*k+1], lazy[k]);
            lazy[2*k+2] = h(lazy[2*k+2], lazy[k]);
        }
        node[k] = g(node[k], r(lazy[k], len));
        lazy[k] = defOp;
    }

    void update(int a, int b, E x, int k=0, int l=0, int r=-1){
        if(r < 0)   r = n;
        eval(k, r-l);
        if(b <= l || r <= a){
            return;
        }else if(a <= l && r <= b){
            lazy[k] = h(lazy[k], x);
            eval(k, r-l);
        }else{
            update(a, b, x, 2*k+1, l, (l+r)/2);
            update(a, b, x, 2*k+2, (l+r)/2, r);
            node[k] = f(node[2*k+1], node[2*k+2]);
        }
    }

    T query(int a, int b, int k=0, int l=0, int r=-1){
        if(r < 0)   r = n;
        eval(k, r-l);
        if(b <= l || r <= a)    return defVal;
        if(a <= l && r <= b)    return node[k];
        T lx = query(a, b, 2*k+1, l, (l+r)/2);
        T rx = query(a, b, 2*k+2, (l+r)/2, r);
        return f(lx, rx);
    }

    T get(int x){
        return node[x+n-1];
    }
};

typedef long long ll;

int main(){
    int n, q;
    cin >> n >> q;
    auto f = [](ll x, ll y)->ll{return min(x,y);};
    auto g = [](ll x, ll y)->ll{return x+y;};
    auto h = [](ll x, ll y)->ll{return x+y;};
    auto r = [](ll x, ll y)->ll{return x;};
    ll defv = 1ll<<60;
    ll defop = 0;
    vector<ll> tmp(n, 0);
    SegL<ll, ll, decltype(f), decltype(g), decltype(h), decltype(r)> seg(tmp, defv, defop, f, g, h, r);
    while(q--){
        int op;
        cin >> op;
        if(op == 0){
            int s, t, x;
            cin >> s >> t >> x;
            seg.update(s, t+1, x);
        }else{
            int s, t;
            cin >> s >> t;
            cout << seg.query(s, t+1) << endl;
        }
    }
    return 0;
}