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>

#define rep(n) for(int i=0;i<n;i++)
#define repp(j, n) for(int j=0;j<n;j++)
#define reppp(i, m, n) for(int i=m;i<n;i++)
#define all(c) c.begin(), c.end()
#define rall(c) c.rbegin(), c.rend()
#define debug(x) cerr << #x << ": " << x << endl

using namespace std;

typedef long long ll;
typedef pair<ll, ll> Pll;
typedef pair<int, int> Pii;

const int MAX_N = (1 << 17);

template<typename T> class LazyRMQ {
    public:
    int n;
    T inf = INT_MAX;
    T data[2*MAX_N-1];
    T lazy[2*MAX_N-1];

    LazyRMQ(int m, T init_value=INT_MAX){
        // 2のべき乗にする
        n = 1;
        while(n < m) n <<= 1;
        rep(2*n-1){
            data[i] = init_value;
            lazy[i] = 0;
        }
    }

    void eval(int k, int kl, int kr){
        if(lazy[k] == 0) return;
        if(data[k] == inf) data[k] = 0;
        data[k] += lazy[k];
        if(kr - kl > 1){
            lazy[2*k+1] += lazy[k];
            lazy[2*k+2] += lazy[k];
        }
        lazy[k] = 0;
    }

    // [s,t)
    void add(int s, int t, int k, int kl, int kr, T x){
        eval(k, kl, kr);
        if(kr <= s || t <= kl) return;
        if(s <= kl && kr <= t){
            lazy[k] += x;
            eval(k, kl, kr);
            // cerr << s << " " << t << " " << k << " " << kl << " " << kr << ": " << data[k] << " " << lazy[k] << endl;
            return;
        }
        int kc = (kl+kr)/2;
        add(s, t, 2*k+1, kl, kc, x);
        add(s, t, 2*k+2, kc, kr, x);
        data[k] = min(data[2*k+1], data[2*k+2]);
    }

    // [s,t)
    T find(int s, int t, int k, int kl, int kr){
        eval(k, kl, kr);
        if(kr <= s || t <= kl) return inf;
        if(s <= kl && kr <= t){
            // cerr << "[" << kl << ", " << kr << ") " << data[k] << endl;
            return data[k];
        }
        int kc = (kl+kr)/2;
        T vl = find(s, t, 2*k+1, kl, kc);
        T vr = find(s, t, 2*k+2, kc, kr);
        // cerr << "[" << kl << ", " << kr << ") " << vl << " " << vr << endl;
        return min(vl, vr);
    }
};


int main(){
    int n,nq,q,s,t,x;
    cin >> n >> nq;
    
    LazyRMQ<int> rmq(n, 0);
    for(int i=0;i<nq;i++){
        cin >> q >> s >> t;
        if(q == 1){
            cout << rmq.find(s, t+1, 0, 0, rmq.n) << endl;
        }else{
            cin >> x;
            rmq.add(s, t+1, 0, 0, rmq.n, x);
        }
    }
}