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;
typedef unsigned long long ull;
typedef pair<ll, ll> P;
typedef pair<ll, P> P3;
typedef pair<P ,P> PP;
constexpr ll MOD = ll(1e9) + 7;
constexpr int IINF = INT_MAX;
constexpr ll LLINF = LLONG_MAX;
constexpr int MAX_N = int(1e5) + 5;
constexpr double EPS = 1e-8;
constexpr int di[] = {0, 1, 0, -1}, dj[] = {1, 0, -1, 0};
#define REP(i, n) for (int i = 0; i < n; i++)
#define REPR(i, n) for (int i = n; i >= 0; i--)
#define SORT(v) sort((v).begin(), (v).end())
#define SORTR(v) sort((v).rbegin(), (v).rend())
#define ALL(v) (v).begin(), (v).end()

template <typename Monoid, typename OperatorMonoid>
struct LazySegmentTree{
private:
    using F = function<Monoid(Monoid, Monoid)>;
    using G = function<Monoid(Monoid, OperatorMonoid)>;
    using H = function<OperatorMonoid(OperatorMonoid, OperatorMonoid)>;
    using P = function<OperatorMonoid(OperatorMonoid, int)>;
    int N;
    vector<Monoid> node;
    vector<OperatorMonoid> lazy;
    F f;
    G g;
    H h;
    P p;
    Monoid e;  // identity element
    OperatorMonoid oe;  // identity element

public:
    LazySegmentTree(){}
    LazySegmentTree(F f, G g, H h, Monoid e, OperatorMonoid oe, P p=[](OperatorMonoid a, int b){return a;}):f(f), g(g), h(h), e(e), oe(oe), p(p){}
    void init(int sz){
        N = 1;
        while(N < sz) N <<= 1;
        node.assign(2*N-1, e);
        lazy.assign(2*N-1, oe);
    }
    void build(vector<Monoid>& v){
        int sz = int(v.size());
        init(sz);
        for(int i=0; i<sz; i++){
            node[i+N-1] = v[i];
        }
        for(int i=N-2; i>=0; i--){
            node[i] = f(node[i*2+1], node[i*2+2]);
        }
    }
    void eval(int k, int len){
        if(lazy[k] != oe){
            node[k] = g(node[k], p(lazy[k], len));
            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]);
            }
            lazy[k] = oe;
        }
    }
    Monoid update(int a, int b, OperatorMonoid x){return update(a, b, x, 0, 0, N);}
    Monoid update(int a, int b, OperatorMonoid x, int k, int l, int r){
        eval(k, r-l);
        if(b <= l || r <= a) return node[k];
        if(a <= l && r <= b){
            lazy[k] = h(lazy[k], x);
            return g(node[k], p(lazy[k], r-l));
        }
        return node[k] = f(update(a,b,x,2*k+1,l,(l+r)/2), update(a,b,x,k*2+2,(l+r)/2,r));
    }
    // [a,b)
    Monoid query(int a, int b){return query(a, b, 0, 0, N);}
    Monoid query(int a, int b, int k, int l, int r){
        eval(k, r-l);
        if(b <= l || r <= a) return e;
        if(a <= l && r <= b) return node[k];
        Monoid vl, vr;
        vl = query(a, b, 2*k+1, l, (l+r)/2);
        vr = query(a, b, 2*k+2, (l+r)/2, r);
        return f(vl, vr);
    }
};

int main() {
    int n, q;
    cin >> n >> q;
    auto f = [=](ll a, ll b){return min(a,b);};
    auto g = [=](ll a, ll b){return a+b;};
    LazySegmentTree<ll, ll> seg(f,g,g,INT_MAX,0);
    vector<ll> v;
    v.assign(n, 0);
    seg.build(v);
    for(int i=0;i<q;i++){
        int c, s, t;
        cin >> c >> s >> t;
        if(c==0){
            ll x;
            cin >> x;
            seg.update(s, t+1, x);
        }
        else{
            cout << seg.query(s,t+1) << endl;
        }
    }
    return 0;
}